123a ppt-all-2

1
CHAPTER 1F Fraction
Whole numbers 1, 2, 3, 4,.. are called natural numbers and
they are used to track whole items.
If x and y are two natural numbers and x can be divided by y,
we say x is a multiple of y or y is a factor of x. Hence 15 is a
multiple of 3 (or 5), and 3 (or 5) is a factor of 15. Note that a
number x is always the multiple of itself and the number 1.
A prime number is a natural number that can only be divided
by 1 and itself. Its factors consist of 1 and itself only.
Hence 2, 3, 5, 7, 11, 13,… are prime numbers.
15 is not a prime number because 15 can be divided by 3 or 5.
The number 1 is not a prime number.
Example A. List the factors and the multiples of 12.
The factors of 12: 1, 2, 3, 4, 6, and 12.
The multiples of 12: 12=1x12, 24=2x12, 36=3x12, 48, 60, etc…
1–F1 The Basics
To factor a number x means to write x as a product, i.e. write
x as a*b for some a and b. For example 12 can be factored as
12 = 2*6 = 3*4 = 2*2*3.
We say the factorization is complete if all the factors are
prime numbers. Hence 12=2*2*3 is factored completely,
2*6 is not complete because 6 is not a prime number.
Exponents
To simplify writing repetitive multiplication, we write 22 for 2*2,
we write 23 for 2*2*2, we write 24 for 2*2*2*2 and so on.
Example B. Calculate.
32 = 3*3 = 9 (note that 3*2 = 6)
43 = 4*4*4 = 64 (note that 4*3 = 12)
We write x*x*x…*x as xN where N is the number of x’s
multiplied to itself. N is called the exponent, or the power of x.
Numbers that are factored completely may be written
using the exponential notation.
Hence, factored completely, 12 = 2*2*3 = 22*3,
200 = 8*25 = 2*2*2*5*5 = 23*52.
Each number has a unique form when its completely factored
into prime factors and arranged in order. This is analogous to
chemistry where each chemical has a unique chemical
composition of basic elements such as H20 for water.
Larger numbers may be factored using a vertical format.
Example C. Factor 144 completely. (Vertical format)
144
12 12
3 4 3 4
2 22 2
Gather all the prime numbers at
the end of the branches we have
144 = 3*3*2*2*2*2 = 3224.
Note that we obtain the same answer
regardless how we factor at each step.

2
Associative Law for Addition and Multiplication
( a + b ) + c = a + ( b + c )
( a * b ) * c = a * ( b * c )
Note: We are to perform the operations inside the “( )” first.
For example, (1 + 2) + 3 = 3 + 3 = 6, which is the same as
1 + (2 + 3) = 1 + 5 = 6.
Subtraction and division are not associative.
For example, ( 3 – 2 ) – 1 = 0 is different from
3 – ( 2 – 1 ) = 2.
Basic Laws
Of the four arithmetic operations +, – , *, and ÷ ,
+, and * behave nicer than – or ÷ . We often take advantage
of the this nice “behavior” of the addition and multiplication
operations to get short cuts. However, be sure you don’t
mistaken apply these short cuts to – and ÷ .
Commutative Law for Addition and Multiplication
a + b = b + a
a * b = b * a
For example, 3*4 = 4*3 = 12.
Subtraction and division don't satisfy commutative laws.
For example: 2  1  1  2.
From the above laws, we get the following important facts.
When adding, the order of addition doesn’t matter.
Hence to add a list numbers, it's easier to add the ones that
add to multiples of 10 first.
ExampleD.
14 + 3 + 16 + 8 + 35 + 15
= 30 + 11 + 50 = 91
When multiplying, the order of multiplication doesn’t
matter.
Example E.
36 = 3*3*3*3*3*3
= 9 * 9 * 9
= 81 * 9 = 729
When multiplying many numbers, always multiply them in
pairs first. This is useful for raising exponents.
Distributive Law : a*(b c) = a*b a*c
Example F.
5*( 3 + 4 ) = 5*7 = 35
or by distribute law,
5*( 3 + 4 ) = 5*3 + 5*4 = 15 + 20 = 35

3
Exercise A.
Do the following problems two ways.
* Add the following by summing the multiples of 10 first.
* Add by adding in the order.
to find the correct answer.
1. 3 + 5 + 7 2. 8 + 6 + 2 3. 1 + 8 + 9
4. 3 + 5 + 15 5. 9 + 14 + 6 6. 22 + 5 + 8
7. 16 + 5 + 4 + 3 8. 4 + 13 + 5 + 7
9. 19 + 7 + 1 + 3 10. 4 + 5 + 17 + 3
11. 23 + 5 + 17 + 3 12. 22 + 5 + 13 + 28
13. 35 + 6 + 15 + 7 + 14 14. 42 + 5 + 18 + 12
15. 21 + 16 + 19 + 7 + 44 16. 53 + 5 + 18 + 27 + 22
17. 155 + 16 + 25 + 7 + 344 18. 428 + 3 + 32 + 227 + 22
Basic Laws
Basic Laws
19. List the all the factors and the first 4 multiples of the
following numbers. 6, 9, 10, 15, 16, 24, 30, 36, 42, 56, 60.
B. Calculate.
1. 33 2. 42 3. 52 4. 53 5. 62 6. 63 7. 72
8. 82 9. 92 10. 102 11. 103 12. 104 13. 105
14. 1002 15. 1003 16. 1004 17. 112 18. 122
20. Factor completely and arrange the factors from smallest to
the largest in the exponential notation: 4, 8, 12, 16, 18, 24,
27, 32, 36, 45, 48, 56, 60, 63, 72, 75, 81, 120.
21. 3 * 5 * 4 * 2 22. 6 * 5 * 4 * 3 23. 6 * 15 * 3 * 2
24. 7 * 5 * 5 * 4 25. 6 * 7 * 4 * 3 26. 9 * 3 * 4 * 4
27. 2 * 25 * 3 * 4 * 2 28. 3 * 2 * 3 * 3 * 2 * 4
29. 3 * 5 * 2 * 5 * 2 * 4 30. 4 * 2 * 3 * 15 * 8 * 4
31. 24 32. 25 33. 26 34. 27 35. 28
36. 29 37. 210 38. 34 39. 35 40. 36
C. Multiply in two ways to find the correct answer.
Fractions are numbers of the form (or p/q) where
p, q  0 are whole numbers. Fractions are numbers that
measure parts of whole items.
Suppose a pizza is cut into 6 equal slices and we have 3 of
them, the fraction that represents this quantity is .
p
q
3
6
The bottom number is the
number of equal parts in the
division and it is called the
denominator.
The top number “3” is the
number of parts that we
have and it is called the
numerator.
3
6
1–F2 Fractions

4
Whole numbers can be viewed as fractions with denominator 1.
Thus 5 = and x = . The fraction = 0, where x  0.
However, does not have any meaning, it is undefined.
5
1
x
1
0
x
x
0
The Ultimate No-No of Mathematics:
The denominator (bottom) of a fraction can't
be 0. (It's undefined if the denominator is 0.)
Fractions that represents the same quantity are called
equivalent fractions.
… are equivalent fractions.
The fraction with the smallest denominator of all the
equivalent fractions is called the reduced fraction.
1
2
=
2
4
=
3
6
=
4
8
is the reduced one in the above list.
1
2
Factor Cancellation Rule
Given a fraction , then
that is, if the numerator and denominator are divided by the
same quantity c, the result will be an equivalent fraction.
In other words, a common factor of the numerator and the
denominator may be canceled as 1, i.e.
a
b
a
b =
a / c
a
b .
=
a*c
b*c
=
a*c
b*c
1
b / c
Example A: Reduce the fraction .78
54
78
54
= 78/2
54/2
= 13
9 .
To reduce a fraction, we keep divide the top and bottom by
common numbers until no more division is possible.
What's left is the reduced version.
=
39/3
27/3
or divide both by 6 in one step.
39
27
(Often we omit writing the 1‟s after the cancellation.)
One common mistake in cancellation is to cancel a common
number that is part of an addition (or subtraction) in the
numerator or denominator.
2 + 1
2 + 3
= 2 + 1
2 + 3
= 1
3
3
5
=
This is addition. Can‟t cancel!
!?
A fraction whose numerator is the same or more than its
denominator (e.g. ) is said to be improper .
We may put an improper fraction into mixed form by division.
Improper Fractions and Mixed Numbers
3
2
2 * 1
2 * 3
=
1
3
Yes
A participant in a sum or a difference is called a term.
The “2” in the expression “2 + 3” is a term (of the expression).
The “2” is in the expression “2 * 3” is called a factor.
Terms may not be cancelled. Only factors may be canceled.

5
23
4
23 4 = 5 with remainder 3. Hence,·
·
23
4
= 5 + 5 3
4 .
5 3
4
= 4*5 + 3
4
23
4=
Improper Fractions and Mixed Numbers
Example B. Put into mixed form.
3
4
=
We may put a mixed number into improper fraction by doing
the reverse via multiplication.
Example C: Put into improper form.5 3
4
Fractions and Mixed Numbers
B. Convert the following improper fractions into mixed
numbers then convert the mixed numbers back to the
improper form.
9
2
11
3
9
4
13
5
37
12
86
11
121
17
1. 2. 3. 4. 5. 6. 7.
Exercise. A. Reduce the following fractions.
4
6 ,
8
12 ,
15
9 ,
24
18 ,
30
42 ,
54
36 ,
60
48 ,
72
108
c
d
=
a*c
b*d
a
b
*
1–F3 Multiplication and Division of Fractions
Example A. Multiply by reducing first.
=
15 * 12
8 * 25
12
25
15
8
*
2
3
5
3
= =
9
10
3*3
2*5
7*8*9*10
8*9*10*11
b.
8
9
7
8
*
10
11
9
10
** = =
7
11
a.
Can't do this for addition and subtraction, i.e.
c
d
=
a c
b d
a
b
±
±
±
Rule for Multiplication of Fractions
To multiply fractions, multiply the numerators and multiply
the denominators, but always cancel as much as possible
first then multiply.

6
a
b
d a
b
d
d
1
Example B: Multiply by cancelling first.
2
3 18 = 2 6 = 12a.
The fractional multiplications are important.
6
11
16
48b.
3
or* *
* *
* = 3 * 11 = 33
Often in these problems the denominator b can be cancelled
against d = .
The often used phrases " (fraction) of .." are translated to
multiplications correspond to this kind of problems.
Example C: a. What is of $108?2
3
* 108 = 2 * 36 = 72 $.
2
3
36
The statement translates into
b. A bag of mixed candy contains 48 pieces of chocolate,
caramel and lemon drops. 1/4 of them are chocolate, 1/3 of
them are caramel. How many pieces of each are there? What
fraction of the candies are lemon drops?
1
3
* 48
16
c. A class has x students, ¾ of them are girls, how many girls
are there?
3
4 * x.
For chocolate, ¼ of 48 is 1
4
* 48 = 12,
12
For caramel, 1/3 of 48 is = 16,
so there are 16 pieces of caramel candies.
The rest 48 – 12 – 16 = 20 are lemon drops. The fraction of
the lemon drops is 20
48 = 20/4
48/4 = 5
12
It translates into multiplication as
so there are 12 pieces of chocolate candies.
The reciprocal (multiplicative inverse) of is .a
b
b
a
Reciprocal and Division of Fractions
Two Important Facts About Reciprocals
I. The product of x with its reciprocal is 1.
So the reciprocal of is ,2
3
3
2 the reciprocal of 5 is ,1
5
and the reciprocal of x is .1
xthe reciprocal of is 3,1
3
2
3
3
2*
II. Dividing by x is the same as multiplying by its reciprocal .
= 1, 5 1
5* = 1, x 1
x* = 1,
1
x
For example, 10 † 2 is the same as 10 , both yield 5.*
1
2
Rule for Division of Fractions
To divide by a fraction x, restate it as multiplying by the
reciprocal 1/x , that is, d
c = a*d
b*c
a
b
*
c
d =a
b †
reciprocate

7
Example D: Divide the following fractions.
15
8
=
12
25
*
8
15
=
12
25 5
3
2
3
9
10
a.
6
9
8
19
8 6
3
2
3
16
†
† = * =b.
6
1 = 30*
1
6 =5d. † 5
Example E: We have ¾ cups of sugar. A cookie recipe calls
for 1/16 cup of sugar for each cookie. How many cookies
can we make?
We can make
3
4 † 1
16 = 3
4 *
16
1 = 3 * 4 = 12 cookies.
4
Multiplication and Division of Fractions
Exercise A. Do the following problems like the example above. Do them mentally.
1. a.
4*
2
3
b.
4*
2
5
c.
4*
2
7
d.
4*
2
9
2. a.
6*
2
3
b.
6*
3
2
c.
6*
2
5
d.
6*
3
4
3. a.
6*
3
5
b.
6*
2
5
c.
6*
6
5
d.
6*
2
9
4. a.
8*
4
3
b.
8*
2
5
c.
8*
4
7
d.
8*
4
5
5. a.
9*
9
3
b.
9*
3
5
c.
12*
4
5
d.
12*
2
9
6. a.
12*
4
3
b.
12*
2
5
c.
12*
6
7
d.
12*
3
4
7. a.
18*
2
3
b.
18*
3
4
c.
18*
6
7
d.
18*
9
7
8. a.
24*
2
3
b.
24*
3
5
c.
24*
4
7
d.
24*
6
5
9. a.
24*
12
7
b.
24*
6
11
c.
36*
4
7
d.
36*
6
5

8
10. a.
36*
2
3
b.
36*
9
5
c.
36*
12
7
d.
36*
3
5
Exercise. B.
11. In a class of 45 people, 2/3 of them are boys, how many boys are there?
12. In a class of 48 people, 1/3 of them are boys, how many girls are there?
13. In a class of 60 people, 3/4 of them are not boys, how many boys are there?
14. In a class of 72 people, 5/6 of them are not girls, how many boys are there?
15. In a class of 56 people, 3/7 of them are not boys, how many girls are there?
16. In a class of 60 people, 1/3 of them are girls, how many are not girls?
17. In a class of 60 people, 2/5 of them are not girls, how are not boys?
18. In a class of 108 people, 5/9 of them are girls, how many are not boys?
A mixed bag of candies has 72 pieces of colored candies, 1/8 of them are red, 1/3 of them are green, ½ of
them are blue and the rest are yellow.
19. How many green ones are there?
20. How many are blue?
21. How many are not yellow?
20. How many are not blue and not green?
21. In a group of 108 people, 4/9 of them adults (aged 18 or over), 1/3 of them are teens (aged from 12 to 17)
and the rest are children. Of the adults 2/3 are females, 3/4 of the teens are males and 1/2 of the children are
girls. Complete the following table.
Adults Teens Children
Male
Female
Total
22. How many females are there and what is the fraction of the females to entire group?
23. How many are not male–adults and what is the fraction of them to entire group?

9
Definition of LCM
The least common multiple (LCM) of two or more numbers is
the least number that is the multiple of all of these numbers.
Example A. Find the LCM of 4 and 6.
The multiples of 4 are 4, 8, 12, 16, 20, 24, …
The multiples of 6 are 6, 12, 18, 24, 30,…
The smallest of the common multiples is 12, so LCM{4, 6 } = 12.
1–F4 LCM and LCD
Example B. Find the LCM of 8, 9, and 12.
The largest number is 12 and the multiples of 12 are 12, 24,
36, 48, 60, 72, 84 … The first number that is also a multiple
of 8 and 9 is 72. Hence LCM{8, 9, 12} = 72.
We may improve the above listing-method for finding the LCM.
Given two or more numbers, we start with the largest number,
list its multiples in order until we find the one that can divide all
the other numbers.
To construct the LCM:
a. Factor each number completely
b. For each prime factor, take the highest power appearing in
the factorizations. The LCM is their product.
Example C. Construct the LCM of {8, 15, 18}.
Factor each number completely,
8 = 23
15 = 3 * 5
18 = 2 * 32
From the factorization select the highest degree of each prime
factor: 23, 32, 5, then LCM{8, 15, 18} = 23*32*5 = 8*9*5 = 360.
But when the LCM is large, the listing method is cumbersome.
It's easier to find the LCM by constructing it instead.
The LCM of the denominators of a list of fractions is called the
least common denominator (LCD). Following is an
application of the LCM.
Example D. From one pizza, Joe wants 1/3, Mary wants 1/4
and Chuck wants 1/6. How many equal slices should we cut
the pizza into and how many slices should each person take?
What is the fractional amount of the pizza they want in total?
Joe gets 12* = 4 slices 1
4
Mary gets 12* = 3 slices
1
6
Chuck gets 12* = 2 slices
We find the LCM of 1/3, 1/4, 1/6 by searching. The multiples of
6 are 6, 12, 18, 24, … Since 12 is also the multiple of 3 and 4,
then 12 is the LCM. Hence we should cut it into 12 slices and
Mary Chuck
In picture:
Joe
1
3
In total, that is 4 + 2 + 3 = 9 slices, or of the pizza.
9
12 = 3
4

10
Your Turn: From one pizza, Joe wants 3/8 of it, Mary wants 1/6
of it and Chuck wants 5/12 of it. How many equal slices should
we cut the pizza and how many slices should each person
take?
In the above example, we found that is the same .
1
3
4
12
The following theorem tells us how to convert the denominator
of a fraction to a fraction with a different denominator.
Multiplier Theorem:
To convert the fraction into a fraction with denominator d,
the new numerator is * d.
a
b
a
b
9
16
Hence =
Example D: Convert to a fraction with denominator 48.
The new denominator is 48, then the new numerator is
48* = 27.
9
16
27
48 .
9
16
3
LCM and LCD
Exercise A. Find the LCM.
1. a.{6, 8} b. {6, 9} c. {3, 4}
d. {4, 10}
2. a.{5, 6, 8} b. {4, 6, 9} c. {3, 4, 5}
d. {4, 6, 10}
3. a.{6, 8, 9} b. {6, 9, 10} c. {4, 9, 10}
d. {6, 8, 10}
4. a.{4, 8, 15} b. {8, 9, 12} c. {6, 9, 15}
5. a.{6, 8, 15} b. {8, 9, 15} c. {6, 9, 16}
6. a.{8, 12, 15} b. { 9, 12, 15} c. { 9, 12, 16}
7. a.{8, 12, 18} b. {8, 12, 20} c. { 12, 15, 16}
8. a.{8, 12, 15, 18} b. {8, 12, 16, 20}
9. a.{8, 15, 18, 20} b. {9, 16, 20, 24}
B. Convert the fractions to fractions with the given
denominators.
10. Convert to denominator 12.
2
3 ,
3
4 ,
5
6 ,
7
4
1
6 ,
3
4 ,
5
6 ,
3
8
7
12 ,
5
4 ,
8
9 ,
11
6
9
10 ,
7
12 ,
13
5 ,
11
15

11
Suppose a pizza is cut into 4 equal slices and Joe takes
one slice or ¼ of the pizza, Mary takes two slices or 2/4 of
the pizza, altogether they take
+
1–F5 Addition and Subtraction of Fractions
Addition and Subtraction of Fractions With the Same
Denominator
To add or subtract fractions of the same denominator, keep
the same denominator, add or subtract the numerators
,then simplify the result.
1
4
2
4 = 3
4
of the entire pizza. In picture:
a
d
b
d = a ± b
d
+ =
1
4
2
4
3
4
Suppose a pizza is cut into 4 equal slices and Joe takes
one slice or ¼ of the pizza, Mary takes two slices or 2/4 of
the pizza, altogether they take
+
Addition and Subtraction of Fractions With the Same
Denominator
To add or subtract fractions of the same denominator, keep
the same denominator, add or subtract the numerators
,then simplify the result.
1
4
2
4 = 3
4
of the entire pizza. In picture:
a
d
b
d = a ± b
d
+ =
1
4
2
4
3
4
Example A:
a. 7
12 + =7 + 11
12
18
12
= 3
2
18/6
12/6 =
+ = 8 + 4 – 2
15
= 2
3=b. 8
15
4
15 – 2
15
10
15
11
12 =
Fractions with different denominators can‟t be added directly
since the “size” of the fractions don‟t match. For example
To add them, first find the LCD of ½ and 1/3, which is 6.
We then cut each pizza into 6 slices. Both fractions may be
converted to have the denominator 6. Specifically,
+
3
6
2
6
=
1
2 = 3
6
1
3 = 2
6 Hence, 1
2 + 1
3 = 3
6 + 2
6 = 5
6
5
6

12
Example A:
a. 7
12
+ =
7 + 11
12
18
12
= 3
2
18/6
12/6
=
+ =
8 + 4 – 2
15
=
2
3
=b.
Addition and Subtraction of Fractions
8
15
4
15
–
2
15
10
15
11
12 =
Fractions with different denominators can‟t be added directly
since the “size” of the fractions don‟t match. For example
To add them, first find the LCD of ½ and 1/3, which is 6.
We then cut each pizza into 6 slices. Both fractions may be
converted to have the denominator 6. Specifically,
+
3
6
2
6
=
1
2
=
3
6
1
3
=
2
6
Hence,
1
2
+
1
3
=
3
6
+
2
6
=
5
6
5
6
We need to convert fractions of different denominators to
a common denominator in order to add or subtract them.
The easiest common denominator to use is the LCD, the least
common denominator. We list the steps below.
Addition and Subtraction of Fractions With the Different
Denominator
1. Find their LCD
2. Convert all the different-denominator-fractions to the have
the LCD as the denominator.
3. Add and subtract the adjusted fractions then simplify the
result.
Example B:
5
6
3
8
+a.
Step 1: To find the LCD, list the multiples of 8 which are
8, 16, 24, .. we see that the LCD is 24.
Step 2: Convert each fraction to have 24 as the denominator.
For , the new numerator is 24 * = 20, hence
5
6
5
6
5
6 =
20
24
For , the new numerator is 24 * = 9, hence
3
8
3
8
3
8 =
9
24
Step 3: Add the converted fractions.
5
6
3
8
+ =
20
24
+
9
24
=
29
24
b.
7
12
5
8
+ The LCD is 48.
= 28=
–
16
9
48
=
28 + 30 – 27
Convert:
7
1248 * so
7
12 =
28
48
= 30
5
848 * so
5
8 =
30
48
= 27
9
16
48 * so
9
16
= 27
48
28
48 +
30
48 –
27
48
=
31
48

13
Example C: a.
The LCD is 24. Multiply the problem by 24, then divide by 24.
5
6
3
8+( ) * 24 / 24 = (4*5 + 3*3) / 24 = 29/24 =
4 3 29
24
We may use the following multiplier-method to add or subtract
fractions to reduce the amount of writing. This method is based
on the fact that if we multiply a quantity x by a, then divide by
a, we get back x.
For example, 2 * 5 / 5 = 10/5 = 2, 3 * 8 / 8 = 24/8 = 3.
Multiplier Method (for adding and subtracting fractions)
To add or subtract fractions, multiply the problem by the LCD
(expand it distributive using law), then divide by the LCD.
5
6
3
8
+
( ) * 48 / 48
= (4*7 + 6*5 – 3*9) / 48
= (28 + 30 – 27) / 48
=
67
12
5
8+ – 16
94 3
The LCD is 48. Multiply the problem by 48, expand the
multiplication, divide the result by 48.
b. 7
12
5
8
+ –
16
9
48
31
Exercise A. Calculate and simplify the answers.
1
2
3
2+1. 2. 3. 4.5
3
1
3+
5
4
3
4+
5
2
3
2+
5
5
3
5–5. 6. 7. 8.6
6
5
6–
9
9
4
9– 1
4
7–
B. Calculate by the Multiplier Method and simplify the
answers.
1
2
1
3+17. 18. 19. 20.1
2
1
3–
2
3
3
2+
3
4
2
5+
5
6
4
7
–21. 22. 23. 24.7
10
2
5–
5
11
3
4
+
5
9
7
15
–
9. 1
2
9– 10. 1
3
8– 11. 4
3
4– 12. 8
3
8–
13. 11
3
5– 14. 9
3
8– 15.14
1
6– 16. 21
3
11–5 6 8

14
C. Addition and Subtraction of Fractions
6
1
4
5
25. 26. 27. 28.
29. 30. 31. 32.
33. 34. 35.
36. 38.37.
39. 40.
6
5
4
7

12
7
9
5

12
5
8
3

16
5
24
7

18
5
12
7

20
3
24
11

15
7
18
5

9
4
6
1
4
3

10
7
6
1
4
5

12
5
6
1
8
3

12
1
9
5
8
7

9
2
16
1
24
5

18
7
12
1
4
5

12
7
16
1
18
5

10
7
18
5
24
7

7
18
443. + 15
7
12
11
+44.
18
19
24
7
– 15
5
4
541. – 6
4
9
5
– 6
3
8
5
+ 42.
12
+
–
5
745. – 12
7
9
1
– 6
5
8
5
+ 46.
8
–3
We start out with a simple mathematics procedure that is often
used in real live. It’s called the digit sum. Just as its name
suggests, we sum all the digits in a number.
The digit sum of 12 is 1 + 2 = 3, is the same as the digit sum
of 21, 111, or 11100. To find the digit sum of
1–F6 Some Facts About Divisibility
add 7 8 9 1 8 2 7 3
15 30
add
45
Hence the digit sum of 78999111 is 45.
If we keep adding the digits, the sums eventually become a
single digit sum – the digit root.
The digit root of 78198273 is 9.
9

15
One simple application of the digit sum is to check if a number
may be divided completely by numbers such as 9, 12, or 18.
I. The Digit Sum Test for Divisibility by 3 and 9.
A bag contains 384 pieces of chocolate, can they be evenly
divided by 12 kids? In mathematics, we ask “is 384 divisible
by 12?” for short. How about 2,349,876,543,214 pieces of
chocolate with 18 kids? Is 2,349,876,543,214 divisible by 18?
If the digit sum or digit root of a number may be divided by 3
(or 9) then the number itself maybe divided by 3 (or 9).
For example 12, 111, 101010 and 300100200111 all have
digit sums that may be divided by 3, therefore all of them may
be divided by 3 evenly. However only 3001002000111, whose
digit sum is 9, may be divided evenly by 9.
Example A. Identify which of the following numbers are
divisible by 3 and which are divisible by 9 by inspection.
a. 2345 b. 356004 c. 6312 d. 870480
II. The Test for Divisibility by 2, 4, and 8
A number is divisible by 2 if its last digit is even.
We refer the above digit–sum check for 3 and 9 as test I.
Hence ****32 is divisible by 4 but ****42 is not.
A number is divisible by 8 if its last 3 digits is divisible by 8.
Hence ****880 is divisible by 8, but ****820 is not.
A number is divisible by 4 if its last 2 digits is divisible by 4 –
you may ignore all the digits in front of them.
III. The Test for Divisibility by 5.
A number is divisible by 5 if its last digit is 5.
From the above checks, we get the following checks for
important numbers such as 6, 12, 15, 18, 36, etc..
The idea is to do multiple checks on any given numbers.
We continue with test II and III.
The Multiple Checks Principle.
If a number passes two different of tests I, II, or III, then it’s
divisible by the product of the numbers tested.
The number 102 is divisible by 3-via the digit sum test.
It is divisible by 2 because the last digit is even.
Hence 102 is divisible by 2*3 = 6 (102 = 6 *17).
Example B. A bag contains 384 pieces of chocolate, can they
be evenly divided by 12 kids? How about 2,349,876,543,214
pieces of chocolate with 18 kids?
For 384 its digit sum is 15 so it’s divisible by 3 (but not 9).
Its last two digits are 84 which is divisible by 4. Hence 384 is
divisible by 3 * 4 or 12.
For 2,349,876,543,210 for 18, since it's divisible by 2, we only
have to test divisibility for 9. Instead of actually find the digit
sum, let’s cross out the digits sum to multiple of 9. We see
that it’s divisible by 9, hence it’s divisible by 18.

16
Exercise. Check each for divisibility by 3 or 6 by inspection.
1. 106 2. 204 3. 402 4. 1134 5. 183420
Check each for divisibility by 4 or 8 by inspection.
6. 116 7. 2040 8. 4020 9. 1096 10. 801171
11. Which numbers in problems 1 – 10 are divisible by 9?
Some Facts About Divisibility
In this section we look at the useful procedure of cross
multiplcation.
What we get are two numbers.
a
b
c
d
Cross Multiplication
Many procedures with two fractions utilize the operation of
cross– multiplication as shown below.
Take the denominators and multiply them diagonally across.
ad bc
Make sure that the denominators cross over and up so the
numerators stay put. Do not cross downward as shown
here. a
b
c
d
adbc
1–F7 Cross Multiplication
Here are some operations where we may cross multiply.
Rephrasing Fractional Ratios
If a cookie recipe calls for 3 cups of sugar and 2 cups of flour,
we say the ratio of sugar to flour is 3 to 2, and it‟s written as
3 : 2 for sugar : flour. Or we said the ratio of flour : sugar is
2 : 3. For most people a recipe that calls for the fractional ratio
of 3/4 cup sugar to 2/3 cup of flour is confusing. It‟s better to
cross multiply to rewrite this ratio in whole numbers.
Example A.
rewrite a recipe that calls for the fractional ratio of 3/4 cup
sugar to 2/3 cup of flour into ratio of whole numbers.
Write 3/4 cup of sugar as and 2/3 cup of flour as3
4 S
2
3 F.
We have the ratio 3
4 S : 2
3 F cross multiply we‟ve 9S : 8F.
Hence in integers, the ratio is 9 : 8 for sugar : flour.
Remark: A ratio such as 8 : 4 should be simplified to 2 : 1.

17
Cross–Multiplication Test for Comparing Two Fractions
Hence cross– multiply
When comparing two fractions to see which is larger and
which is smaller. Cross–multiply them, the side with the larger
product corresponds to the larger fraction.
In particular, if the cross multiplication products are the same
then the fraction are the same.
3
5
9
15
=45 45 so
3
5
9
15
=
we get
Cross– multiply 3
5
5
8
24 25
Hence 3
5
5
8
is less than
we get
moreless
.
(Which is more 7
11
9
14 or ? Do it by inspection.)
Cross–Multiplication Test for Comparing Two Fractions
Hence cross– multiply
When comparing two fractions to see which is larger and
which is smaller. Cross–multiply them, the side with the larger
product corresponds to the larger fraction.
In particular, if the cross multiplication products are the same
then the fraction are the same.
3
5
9
15
=45 45 so
3
5
9
15
=
we get
Cross– multiply 3
5
5
8
24 25
Hence 3
5
5
8
is less than
we get
moreless
.
(Which is more 7
11
9
14 or ? Do it by inspection.)
Cross–Multiplication for Addition or Subtraction
We may cross multiply to add or subtract two fractions with
the product of the denominators as the common denominator.
a
b
c
d
Afterwards we reduce if necessary for the simplified answer.
Example B. Calculate
± = ad ± bc
bd
3
5
5
6
– =
5*5 – 6*3
6*5
7
30=a.
5
12
5
9
– =5*12 – 9*5
9*12
15
108=b. 5
36=
Cross multiplication is the same as clearing the two
denominators by using their product. We may use their LCM
to clear the denominators but the LCM requires extra work.

18
Cross–Multiplication for Addition or Subtraction
We may cross multiply to add or subtract two fractions with
the product of the denominators as the common denominator.
a
b
c
d
Afterwards we reduce if necessary for the simplified answer.
Example B. Calculate
± = ad ± bc
bd
3
5
5
6 – = 5*5 – 6*3
6*5
7
30
=a.
5
12
5
9 – =5*12 – 9*5
9*12
15
108
=b. 5
36
=
Cross multiplication is the same as clearing the two
denominators by using their product. We may use their LCM
to clear the denominators but the LCM requires extra work.
Exercise A. Restate the following ratios in integers.
9. In a market, ¾ of an apple may be traded with ½ a pear.
Restate this using integers.
1
2
1
3
:1. 2. 3. 4.2
3
1
2
:
3
4
1
3
:
2
3
3
4
:
3
5
1
2
:5. 6. 7. 8.1
6
1
7
:
3
5
4
7
:
5
2
7
4
:
B. Determine which fraction is more and which is less.
2
3
3
4
,10. 11. 12. 13.4
5
3
4
,
4
7
3
5
,
5
6
4
5
,
5
9
4
7
,14. 15. 16. 17.7
10
2
3
,
5
12
3
7
,
13
8
8
5
,
C. Use cross–multiplication to combine the fractions.
1
2
1
3
+18. 19. 20. 21.1
2
1
3
–
2
3
3
2
+
3
4
2
5
+
5
6
4
7
–22. 23. 24. 25.7
10
2
5
–
5
11
3
4
+
5
9
7
15
–
Cross Multiplication

19
CHAPTER 1B Signed Numbers
1–S1 Signed Numbers
Example A:
a. We deposited $400 into a bank account then withdrew $350
from the account, write the transactions using signed numbers.
How much is left in the account?
We use “+” for deposit or having surplus in the account
and “–” for withdraw or debit from the account, then the
transactions may be listed as: +400, –350. (In this section we
will use red color for negative numbers for emphasis.)
The amount left in the account is $50. This is a surplus so
its +50. We write the entire transactions as +400 – 350 = +50.
We track the “directions” of measurements by giving them
positive (+) or negative (-) sign. Signed measurements
represent amounts of increases versus decreases, surpluses
versus deficiencies, credits versus debits and so on. Numbers
with signs are called signed numbers.
b. We deposited $400 into the account then withdrew $600
from the account, write the transactions using signed numbers.
How much is left in the account?
The transactions may be listed as: +400, –600.
The account is short by $200. This is a deficiency so its –200.
We write the entire transactions as +400 – 600 = –200
c. We deposited $400 in a bank account, later deposited
another $200, then withdrew $350, then deposited $250, then
withdrew $600, make a list of the transactions using signed
numbers. How much is left?
The transactions are listed as +400, +200, –350, +250, –600.
To find the final balance in the account, we note that the total
of the deposits is +850 and the total of the withdrawals is –950,
so the account is short of $100, or there is “–100” left in the
account. We write these transactions as
+400 + 200 – 350 + 250 – 600 = –100.
Example B:
+100 + 200 = +300, –100 – 200 = –300
+500 – 300 = +200, –500 + 300 = –200
A number written without a sign is treated as a positive
number. Therefore, 100 + 200 is the same as +100 + 200
and both combined to 300.
In order to state precisely the rules for combining signed
numbers, we introduce the notion of absolute values.
In example A of the bank account, if we are only interested in
the amount of the transactions but not the type of transactions,
this amount is called the absolute value.
The absolute value of a number x is written as |x|.
Hence, |500| = 500, |-350| = 350, |-600| = 600, etc..
The above operation of totaling two or more signed numbers
into a single signed number is called the combining operation.

20
Rules for Combining Signed Numbers
I. To combine two or more numbers of the same signs, keep
the sign, take the sum of the absolute values of the numbers.
+100 + 200 = +300, –100 – 200 = –300
II. To combine two numbers of different signs, keep the sign
of the number with larger absolute value, take the difference
of the absolute values of the numbers.
+500 – 300 = +200, –500 + 300 = –200
Example C: 8 – 9 + 11 = –1 + 11 = 10
To combine many numbers, an alternative way is to do it as in
example A of the bank transactions. That is, we combined all
the positive ones (deposits) first, then combine all the negative
ones (withdrawals), then combine the two results.
There are different ways to combine multiple signed numbers,
we may combine them from left to right.
Exercise A. Combine
1. 2 + 3 2. 10 + 6 3. 34 + 21 + 4 + 17 4. –6 –2
5. –11 – 5 6. –14 –15 7. –26 –15 – 5 –14
8. –3 + 2 9. 5 –11 10. –14 + 15
11. 26 –15 12. 12 – 13 13. –23 +18
B. Combine by moving the positive numbers to the front first.
Combine the positive numbers, the negative numbers
separately then then combine the two results.
14. 23 – 18 +7 –12 15. –6 –2 + 10 + 6
16. –14 + 23 –15 – 3 +12 17. –26 + 15 –5 –14 + 9
18. 19 – 13 – 9 – 3 + 15 19. –6 + 19 – 15 + 5 – 9
20. – 4 + 7 – 23 + 8 + 17 – 8 + 6 + 9 – 22 – 2
21. Try to get the same answer for #20 by combining two
numbers at a time without separating the positive
numbers from the negative numbers.
Signed Numbers

21
Addition of Signed Numbers
Adding signed numbers is the same as combining the numbers.
Rule for Addition of Signed Numbers:
To add two signed numbers, remove the parenthesis and
combine the numbers, that is,
a  (+b) = a + b , a + (–b) = a – b
1–S2 Addition and Subtraction of Signed Numbers
Example A. Remove parentheses then combine.
a. 5  (+4)  remove “( )”
= 5 + 4 = 9
b. –7  (3)  remove “( )”
= –7 + 3 = – 4
c. 2  (–6)  remove “( )”
= 2 – 6 = – 4
d. –4  (–8)  remove “( )”
= – 4 – 8 = –12
e. 5  (4) + (–3) + (–8) + 4 – 6 remove ( )‟s
= 5 + 4 – 3 – 8 + 4 – 6
= 5 + 4 + 4 – 3 – 8 – 6
= 13 – 17 = –4
Subtraction of Signed Numbers
For subtraction of signed numbers, we need the notion of
“opposite” numbers. The numbers x and –x are said to be the
opposite or the negative of each other.
The opposite of x is –x. The opposite of –x is –(–x) = x.
So the opposite of 6 is –6 , the opposite of –12 is –(–12) = 12.
Note that the opposite of a negative number is positive.
Rule for Subtraction of Signed Numbers:
To subtract a signed number x, remove the parenthesis and
combine with the opposite of x, that is,
a – (+b) = a – b a – (–b) = a + b
opposite opposite
Example B. Remove parentheses then combine.
a. 5 – (+4)  remove “( )”, change to opposite
= 5 – 4 = 1
b. 3 – (–7)  remove “( )”, change to opposite
= 3 + 7 = 10
c. –12 – (–5)  remove “( )”, change to opposite
= –12 + 5 = –7
Summary for removing parentheses for addition and
subtraction of signed numbers:
For addition: a  (+b) = a + b a + (–b) = a – b
For subtraction: a – (+b) = a – b a – (–b) = a + b
Example C. Remove the parentheses, then combine.
a. –6 – (–8) – (–2) – (9) remove “( )”
= –6 + 8 + 2 – 9
= 8 + 2 – 6 – 9 = 10 – 15 = –5

22
b. 2  (–4) – (–8) – (5)  (–9 ) remove the ( )‟s
= 2 – 4 + 8 – 5 – 9
= 2 + 8 – 4 – 5 – 9
= 10 – 18
= –8
The ( ), [ ], or { } are grouping symbols. Each set of symbol
must contain both the left-hand and right-hand parts. Each set
encloses calculations that are to be done within the symbol.
Example D.
a. –6 – (8 – 9) do the calculation inside the “( )”
= – 6 – (– 1) remove parentheses
= – 6 + 1
= – 5
b. (–6 – 8) – 9 do the calculation inside the “( )”
= (–14) – 9 remove parentheses
= – 14 – 9 = – 23
Example E.
2 – [–6 – (8 + 9)] do the calculation inside the “( )”
= 2 – [–6 – (17)] remove parentheses
= 2 – [– 6 – 17] do the calculation inside the “[ ]”
= 2 – [– 23]
= 2 + 23
= 25
If there is a set of grouping symbol inside another set of
grouping symbol, the inner set is to be calculated first.
Exercise A. Drop the parentheses and write the expressions in
a simpler form, then combine.
1. 7  (+3) 2. 7  (–3) 3. –7  (3)
4. –7  (– 3) 5. –7  (3) + 4 6. –7  (–3) – 4
7. –17  (– 23) + 24 8. 16  (–31) – 22
9. –8  (–15) – 9
B. Drop the parentheses after the subtraction (don‟t forget to
switch the sign) and write the expressions in a simpler form,
then combine.
11. 7 – (+3) 12. 7 – (–3) 13. –7 – (3)
14. –7 – (– 3) 15. –7 – (3) + 4 16. –7 – (–3) – 4
17. –17 – (– 23) + 24 18. 16 – (+31) – 22
19. –8 – (–15) – 9
10. –8  (–15) – 9 + (– 3)
20. –8 – (–15) –(– 19) – (+42) – 3

23
C. Drop the parentheses after the arithmetic operations and
write the expressions in a simpler form, then combine.
21. 7 – (+3) + (– 3) 22. 7 – (–3) + (3)
23. –7 – (– 3) + (– 5) 24. –8 + (– 6) – (–3) – (5)
25. 17 – (– 6) + 4 – (–12) – 11
27. 5 – (–13) – 41 – (–32) – 18 – (–41)
26. –17 – (– 23) + 4 + (–16) – 11
28. –8 – (–15) – 9 + 35 – 7 + (– 25) – 10
D. Work the problems starting from the inside. Make sure that
you copy the ( )‟s, [ ]‟s correctly.
29. 7 – [+3 + (– 3)] 30. [7 – (–3)] + (3)
31. –7 – [(–3) + (– 5)] 32. –8 + [(– 6) – (–3)] – 5
35. 5 – [–13 – 41] – [(–32) – 18] – (–41)
34. –[17 – (– 23)] – [(–16) – 11]
36. –[8 – (–15)] – [9 + 35] – [7 + (– 25) – 10]
33. –8 + (– 6) – [(–3) – 5]
Addition and Subtraction of Signed Numbers
Rule for Multiplication of Signed Numbers
To multiple two signed numbers, we multiply their absolute
values and use the following rules for the sign of the product.
+ * + = – * – = + ;
+ * – = – * + = – ;
1–S3 Multiplication and Division of Signed Numbers
Two numbers with the same sign multiplied yield a positive
product.
Two numbers with opposite signs multiplied yield a negative
product.
Example A.
a. 5 * (4) = –5 * (–4) = 20
b. –5 * (4) = 5 * (–4) = –20
In algebra, multiplication operation are not always written
down explicitly. Instead we use the following rules to identify
multiplication operations.
However, if there is a “+” or “–” sign between the ( ) and a
quantity, then the operation is to combine.
Hence 3(+5) = (+5)3 =15, but 3 + (5) = (3) + 5 = 8,
and –5(–5) = (–5)(–5) = 25, but (–5) – 5 = –5 – (5) = –10.
● If there is no operation indicated between two sets of ( )‟s,
the operation between them is multiplication.
Hence (x + y)(a + b) = (x + y) * (a + b)
● If there is no operation indicated between a set of ( ) and a
quantity, the operation between them is multiplication.
Hence x(a + b) = x * (a + b ).
When multiplying many signed numbers at once, find the sign
of the product first by the following rules.
● If there is no operation indicated between two quantities, the
operation between them is multiplication. Hence xy means x * y.

24
The General Sign Rule (for * and /)
• If there are even number of negative numbers in the
multiplication, the product is positive.
• If there are odd number of negative numbers in the
multiplication, the product is negative.
Example B.
a. –1(–2 ) 2 (–1) = – 4
b. (–2)4 = (–2 )(–2)(–2)(–2) = 16
Fact: A quantity raised to an even power i.e. xeven
is always positive (or 0).
three negative numbers, so the product is negative
four negative numbers, so the product is positive
General Rule for the Sign of a Product
• If there are even number of negative numbers in the
multiplication, the product is positive.
• If there are odd number of negative numbers in the
multiplication, the product is negative.
Example B.
a. –1(–2 ) 2 (–1) = – 4
b. (–2)4 = (–2 )(–2)(–2)(–2) = 16
Fact: A quantity raised to an even power i.e. xeven
is always positive (or 0).
three negative numbers, so the product is negative
four negative numbers, so the product is positive
Make sure that you interpret the operations correctly.
Exercise A. Calculate the following expressions.
1. 3 – 3 2. 3(–3) 3. (3) – 3 4. (–3) – 3
5. –3(–3) 6. –(–3)(–3) 7. (–3) – (–3) 8. –(–3) – (–3)
B.Multiply. Determine the sign first.
9. 2(–3) 10. (–2)(–3) 11. (–1)(–2)(–3)
12. 2(–2)(–3) 13. (–2)(–2)(–2) 14. (–2)(–2)(–2)(–2)
15. (–1)(–2)(–2)(–2)(–2) 16. 2(–1)(3)(–1)(–2)
17. 12
–3
18. –12
–3
19. –24
–8
21. (2)(–6)
–8
C. Simplify. Determine the sign and cancel first.
20. 24
–12
22. (–18)(–6)
–9
23. (–9)(6)
(12)(–3)
24. (15)(–4)
(–8)(–10)
25. (–12)(–9)
(– 27)(15)
26. (–2)(–6)(–1)
(2)(–3)(–2)
27. 3(–5)(–4)
(–2)(–1)(–2)
28. (–2)(3)(–4)5(–6)
(–3)(4)(–5)6(–7)
Multiplication and Division of Signed Numbers

25
If we have two $5-bill and two $10-bills, we have the total of
2(5) + 2(10) = 30 dollars. To get the correct answer 30,
we multiply the 2 and the 5 and multiply the 2 and the10 first,
then we add the products 10 and 20.
If I have three $10-bills and you have four $10-bills, we have
3 + 4 = 7 $10-bills, and we have a total of (3 + 4)10 = 70 $.
In this case, we group the 3 + 4 in the “( )” to indicate that we
are to add them first, then multiply the sum to 10.
1–S4 Order of Operations
Order of Operations (excluding raising power)
Given an arithmetic expression, we perform the operations in
the following order .
1st . Do the operations within grouping symbols, starting with
the innermost grouping symbol.
2nd. Do multiplications and divisions (from left to right).
3rd. Do additions and subtractions (from left to right).
This motivates us to set the rules for the order of operations.
Example A.
a. 4(–8) + 3(5)
= –32 + 15
= –17
c. 9 – 2[7 – 3(6 + 1)]
= 9 – 2[7 – 3(7)]
= 9 – 2[7 – 21]
= 9 – 2[ –14 ]
= 9 + 28
= 37
(Don‟t perform “4 + 3” or “9 – 2” in the above problems!!)
b. 4 + 3(5 + 2)
= 4 + 3(7)
= 4 + 21
= 25
Do now. Don’t do the part that you shouldn’t do!
1. 6 + 3(3 + 1) 2. 10 – 4(2 – 4)
3. 5 + 2[3 + 2(1 + 2)] 4. 5 – 2[3 + 2(5 – 9)]
Ans: a. 18 b. 18 c. 23 4. 15
Exponents
We write x*x*x…*x as xN where N is the number of copies of
x‟s multiplied to itself. N is called the exponent, or the power
of x, and x is called the base.
The base is the quantity immediately beneath the exponent,
hence 2b3 means 2*b3 = 2*b*b*b.
If we want multiply 2b to itself three times, i.e. 2b to the third
power, we write it as (2b)3 which is (2b)*(2b)*(2b) =8b3.

26
c. Expand (3*2)2 and simplify the answer.
The base for the 2nd power is (3*2).
Hence(3*2)2 is (3*2)(3*2) = (6)(6) = 36
d. Expand 3*22 and simplify the answer.
The base for the 2nd power is 2.
Hence 3*22 means 3*2*2 = 12
b. Expand – 32
The base of the 2nd power is 3.
Hence – 32 means – (3*3) = – 9
Example B. (Exponential Notation)
a. Expand (–3)2 and simplify the answer.
The base is (–3).
Hence (–3)2 is (–3)(–3) = 9
e. Expand (–3y)3 and simplify the answer.
(–3y)3
= (–3y)(–3y)(–3y) (the product of three negatives number is negative)
= –(3)(3)(3)(y)(y)(y)
= –27y3
Order of Operations (PEMDAS)
1st. (Parenthesis) Do the operations within grouping symbols,
starting with the innermost one.
2nd. (Exponents) Do the exponentiation
3rd. (Multiplication and Division) Do multiplications and
divisions in order from left to right.
4th. (Addition and Subtraction) Do additions and
subtractions in order from left to right.
From part b above, we see that the power is to be carried out
before multiplication. Below is the complete rules of order of
operations.
e. Expand (–3y)3 and simplify the answer.
(–3y)3
= (–3y)(–3y)(–3y) (the product of three negatives number is negative)
= –(3)(3)(3)(y)(y)(y)
= –27y3
Order of Operations (PEMDAS)
1st. (Parenthesis) Do the operations within grouping symbols,
starting with the innermost one.
2nd. (Exponents) Do the exponentiation
3rd. (Multiplication and Division) Do multiplications and
divisions in order from left to right.
4th. (Addition and Subtraction) Do additions and
subtractions in order from left to right.
From part b above, we see that the power is to be carried out
before multiplication. Below is the complete rules of order of
operations.

27
Example C. Order of Operations
a. 52 – 32
= 25 – 9
= 16
b. – (5 – 3)2
= – (2)2
= – 4
c. –2*32 + (2*3)2
= –2*9 + (6)2
= –18 + 36
= 18
d. –32 – 5(3 – 6)2
= –9 – 5(–3)2
= –9 – 5(9)
= –9 – 45 = –54
Order of Operations
Make sure that you interpret the operations correctly.
Exercise A. Calculate the following expressions.
Order of Operations
7. 1 + 2(3) 8. 4 – 5(6) 9. 7 – 8(–9)
1. 3(–3) 2. (3) – 3 3. 3 – 3(3) 4. 3(–3) + 3
5. +3(–3)(+3) 6. 3 + (–3)(+3)
B. Make sure that you don‟t do the ± too early.
10. 1 + 2(3 – 4) 11. 5 – 6(7 – 8) 12. (4 – 3)2 + 1
13. [1 – 2(3 – 4)] – 2 14. 6 + [5 + 6(7 – 8)](+5)
15. 1 + 2[1 – 2(3 + 4)] 16. 5 – 6[5 – 6(7 – 8)]
17. 1 – 2[1 – 2(3 – 4)] 18. 5 + 6[5 + 6(7 – 8)]
19. (1 + 2)[1 – 2(3 + 4)] 20. (5 – 6)[5 – 6(7 – 8)]
C. Calculate. Make sure apply the powers to the correct bases.
23. (–2)2 and –22 24 (–2)3 and –23 25. (–2)4 and –24
26. (–2)5 and –25 27. 2*32 28. (2*3)2
21. 1 – 2(–3)(–4) 22. (–5)(–6) – (–7)(–8)
D.Make sure that you apply the powers to the correct bases.
29. (2)2 – 3(2) + 1 30. 3(–2)2 + 4(–2) – 1
31. –2(3)2 + 3(3) – 5 32. –3(–1)2 + 4(–1) – 4
33. 3(–2)3 – 4(–2)2 – 1 34. (2)3 – 3(2)2 + 4(2) – 1
35. 2(–1)3 – 3(–1)2 + 4(–1) – 1 36. –3(–2)3 – 4(–2)2 – 4(–2) – 3
37. (6 + 3)2 38. 62 + 32 39. (–4 + 2)3 40. (–4)3 + (2)3
E. Calculate.
41. 72 – 42 42. (7 + 4)(7 – 4 )
43. (– 5)2 – 32 44. (–5 + 3)(–5 – 3 )
45. 53 – 33 46. (5 – 3) (52 + 5*3 + 32)
47. 43 + 23 48. (4 + 2)(42 – 4*2 + 22)
7 – (–5)
5 – 353.
8 – 2
–6 – (–2)54.
49. (3)2 – 4(2)(3) 50. (3)2 – 4(1)(– 4)
51. (–3)2 – 4(–2)(3) 52. (–2)2 – 4(–1)(– 4)
(–4) – (–8)
(–5) – 3
55.
(–7) – (–2)
(–3) – (–6)
56.

28
In mathematics we use symbols such as x, y and z to
represent numbers. These symbols are called variables
because their values change depending on the situation .
We use variables and mathematics operations to make
expressions which are calculation procedures.
For example, if an apple cost $2 and x represents the number
of apples, then “2x” is the expression for the cost for x apples.
Suppose we have 6 apples, set x = 6 in the expression 2x,
we obtain 2(6) = 12 for the total cost.
The value “6” for x is called input (value). The answer 12 is
called the output. This process of replacing the variables with
input value(s) and find the output is called evaluation.
1–S5 Variables and Evaluation
Each variable can represent one specific measurement only.
Suppose we need an expression for the total cost of apples
and pears and x represents the number of apples, we must
use a different letter, say y, to represent the number of pears
since they are two distinct measurements.
Example A.
a. Evaluate –x if x = –6.
When evaluating, insert the input enclosed in a “( )”.
Therefore, set x = (–6) we‟ve
–x  – (–6) = 6
b. Evaluate –3x if x = –6.
–3x  –3(–6) = 18
c. Evaluate –2x2 if x = 6.
–2x2  –2(6)2 = –2(36) = –72
When evaluating an expression, replace the variables with the
input-values enclosed with ( )‟s.
d. Evaluate –4xyz if x = –3, y = –2, z = –1.
–4xyz
–4(–3)(–2)(–1) = 24
f. Evaluate 3x2 – y2 if x = 2 and y = –3.
Replace x by (2) and y by (–3) in the expression, we have
3*(2)2 – (–3)2
= 3*4 – 9
= 12 – 9
= 3
e. Evaluate x – y if x = –3, y = –5.
x – y  (–3) – (–5) = –3 + 5 = 2
g. Evaluate –x2 + (–8 – y)2 if x = 3 and y = –2.
Replace x by 3, y by (–2) in the expression,
– (3)2 + (–8 – (– 2))2
= – 9 + (–8 + 2)2
= – 9 + (–6)2
= – 9 + 36
= 27

29
h. Evaluate (a – b)(b – c) if a = 3, b = –2, c = –4.
(a – b)(b – c)
((3) – (–2))((–2) – (–4))
= (3 + 2)(–2 + 4)
= (5)(2)
= 10
j. Evaluate b2 – 4ac if a = –2, b = –3, and c = 5.
(–3)2 – 4(–2)(5)
= 9 + 40
i. Evaluate (2b – 3a)2 if a = –4, b = – 3.
(2(–3) –3(–4))2
= (–6 + 12)2
= (6)2 = 36
Exercise. Evaluate.
A. –2x with the input
Variables and Evaluation
1. x = 3 2. x = –3 3. x = –5 4. x = –1/2
B. –y – 2x with the input
5. x = 3, y = 2 6. x = –2, y = 3
7. x = –1, y = –4 8. x = ½, y = –6
C. (–x)2 with the input
9. x = 3 10. x = –3 11. x = –5 12. x = –1/2
D. –x2 with the input
13. x = –2 14. x = –2 15. x = –9 16. x = –1/3
E. –2x3 with the input
17. x = 3 18. x = –2 19. x = –1 20. x = –½
F. 3x2 – 2x – 1 with the input
21. x = – 4 22. x = –2 23. x = –1 24. x = ½
G. –2y2 + 3x2 with the input
25. x = 3, y = 2 26. x = –2, y = – 3
27. x = –1, y = –4 28. x = –1, y = –1/2
J. b2 – 4ac with the input
37. a = –2, b = 3, c = –5 38. a = 4, b = –2, c = – 2
39. a = –1, b = – 2, c = –3 40. a = 5, b = –4, c = 4
H. x3 – 2x2 + 2x – 1 with the input
29. x = 1 30. x = –1 31. x = 2 32. x = ½
33. a = –1, b = – 2 34. a = 2, b = –4
–b
2a
I. with the input
35. a = –2, b = – 8 36. a = 2, b = – 12

30
a – b
c – d
K. with the input
43. a = –2, b = 3, c = –5, d = 0
44. a = –1, b = –2, c = –2, d = 14
41. a = 1, b = –2, c = 2, d = – 2
42. a = –4, b = –2, c = –1, d = –4
(a – b)(b – c)
(c – d)(d – a)
L. with the input
47. a = –2, b = 3, c = –5, d = 0
48. a = –1, b = –2, c = –2, d = 14
45. a = 1, b = –2, c = 2, d = – 2
46. a = –4, b = –2, c = –1, d = –4
M. b2 – a2 – c2 if
49. a = –2, b = 3, c = –5 .
50. a = 4, b = –2, c = – 2

31
Chapter 2 Linear Equations and Inequalities
Example A.
2 + 3x  “the sum of 2 and 3 times x”
4x2 – 5x  “the difference between 4 times the square of x
and 5 times x”
(3 – 2x)2  “the square of the difference between 3 and twice
x”
A mathematical expression is a calculation procedure written
using numbers, variables, and operation symbols.
2–1 Expressions
The simplest type of expressions are linear expressions.
These are expressions of the form ax + b where the a and b
are numbers.
Example B.
3x – 4, 6 + 2x, –5 are linear expressions,
2x2 – 3, are not linear.
1
x
For the x-term ax, the number “a” is called the coefficient of
the term. When we multiply a number with an x-term, we
multiply it with the coefficient. Hence, 3(5x) = (3*5)x =15x,
and –2(–4x) = (–2)(–4)x = 8x.
We may multiply a number with an expression and expand the
result by the distributive law.
Distributive Law
A(B ±C) = AB ±AC = (B ±C)A
Example C. Combine.
2x – 4 + 9 – 5x
= 2x – 5x – 4 + 9
= –3x + 5
Example D. Expand then simplify.
a. –5(2x – 4)
= –5(2x) – (–5)(4)
= –10x + 20
Just as 2 apples + 3 apples = 5 apples we may combine two
x-terms. Hence 2x + 3x = 5x.
We combine positive x-terms and negative x-terms the same
way we combine signed numbers. Hence –3x – 5x = –8x.
The number-terms may be combined such as 3 – 8 = –5.
However just as 2 apple + 3 banana = 2 apple + 3 banana,
(i.e. they can‟t be combined), x-terms may not be combined
with number terms because they are different type of items.
Hence 2 + 3x stays as 2 + 3x (not 5x).
Combining Expressions
Given an expression, each addend is called a term. There are
two terms in ax + b.
the x-term the number term or the constant term

32
b. 3(2x – 4) + 2(4 – 5x)
= 6x – 12 + 8 – 10x
= –4x – 4
Distributive law gives us the option of expanding the content of
the parentheses i.e. extract items out of containers.
Let A stands for apple and B stands for banana,
then Regular = (12A + 8B) and Deluxe = (24A + 24B).
Three boxes of Regular and four boxes of Deluxe is
3(12A + 8B) + 4(24A + 24B)
= 36A + 24B + 96A + 96B = 132A + 120B
Hence we have 132 apples and 120 bananas.
Example E. A store sells two types of gift boxes Regular and
Deluxe. The Regular has 12 apples and 8 bananas, the Deluxe
has 24 apples and 24 bananas. We have 3 boxes of Regular and
4 boxes of Deluxe. How many apples and bananas are there?
We usually start with the innermost set of parentheses to
simplify an expression that has multiple layers of parentheses.
In mathematics, ( )‟s, [ ]‟s, and { }‟s are often used to
distinguish the layers. This may not be the case for calculators
or software where [ ] and { } may have other meanings.
Always simplify the content of a set of parentheses first before
expanding it.
= –3{–3x – [5 + 8x + 12] – 4}
= –3{–3x – [17 + 8x] – 4}
Example F. Expand.
= –3{– 3x – 17 – 8x – 4}
–3{–3x – [5 – 2(– 4x – 6)] – 4} expand,
simplify,
expand,
simplify,
= –3{–11x – 21} expand,
= –33x + 63
X-terms with fractional coefficients may be written in two ways,
p
q
x or
px
q
( but not as ) since
p
qx
p
q x =
p
q
x
1 = px
q
Example G. Evaluate if x = 6.4
3
x
we getSet x = (6) in 4
3
2
Hence is the same as .2x
3
2
3 x
*
4
3
x, (6) = 8
We may use the multiplier method to combine fraction terms
i.e. multiply the problem by the LCD and divide by the LCD.
31x
12
4 3
= ( )12 / 12
4
3
x + 5
4
x
4
3
x + 5
4
x expand and cancel the denominators,
= (4*4 x + 5*3x) / 12 = (16x + 15x) /12 =
Example H. Combine
4
3
x + 5
4
x
Multiply and divide by their LCD =12,

33
42. As in example C with gift boxes Regular and Deluxe, the
Regular contains12 apples and 8 bananas, the Deluxe has 24
apples and 24 bananas. For large orders we may ship them in
crates or freight-containers where a crate contains 100 boxes
Regular and 80 boxes Deluxe and a container holds 150
Regular boxes and 100 Deluxe boxes.
King Kong ordered 4 crates and 5 containers, how many of
each type of fruit does King Kong have?
41. As in example C with gift boxes Regular and Deluxe, the
Regular contains12 apples and 8 bananas, the Deluxe has 24
apples and 24 bananas. Joe has 6 Regular boxes and 8
Deluxe boxes. How many of each type of fruit does he have?
C. Starting from the innermost ( ) expand and simplify.
29. x + 2[6 + 4(–3 + 5x)] 30. –5[ x – 4(–7 – 5x)] + 6
31. 8 – 2[4(–3x + 5) + 6x] + x 32. –14x + 5[x – 4(–5x + 15)]
33. –7x + 3{8 – [6(x – 2) –3] – 5x}
34. –3{8 – [6(x – 2) –3] – 5x} – 5[x – 3(–5x + 4)]
35. 4[5(3 – 2x) – 6x] – 3{x – 2[x – 3(–5x + 4)]}
2
3
x + 3
4
x36.
4
3
x – 3
4
x37.
3
8
x – 5
6
x39.5
8
x + 1
6
x38. – –
D. Combine using the LCD-multiplication method
40. Do 36– 39 by the cross–multiplication method.
Exercise A. Combine like terms and simplify the expressions.
Expressions
1. 3x + 5x 2. 3x – 5x 3. –3x – 5x 4. –3x + 5x
5. 3x + 5x + 4 6. 3x – 5 + 2x 7. 8 – 3x – 5
8. 8 – 3x – 5 – x 9. 8x – 4x – 5 – 2x 10. 6 – 4x – 5x – 2
11. 3A + 4B – 5A + 2B 12. –8B + 4A – 9A – B
B. Expand then simplify the expressions.
13. 3(x + 5) 14. –3(x – 5) 15. –4(–3x – 5) 16. –3(6 + 5x)
25. 3(A + 4B) – 5(A + 2B) 26. –8(B + 4A) + 9(2A – B)
17. 3(x + 5) + 3(x – 5) 18. 3(x + 5) – 4(–3x – 5)
19. –9(x – 6) + 4(–3 + 5x) 20. –12(4x + 5) – 4(–7 – 5x)
21. 7(8 – 6x) + 4(–3x + 5) 22. 2(–14x + 5) – 4(–7x – 5)
23. –7(–8 – 6x) – 4(–3 – 5x) 24. –2(–14x – 5) – 6(–9x – 2)
27. 11(A – 4B) – 2(A – 12B) 28. –6(B – 7A) – 8(A – 4B)

34
We want to solve equations. That is, we want to find a value
(or values) for the variable x such that it makes the two sides
equal. Such a value is called a solution of the equation.
Example A.
For 14 = 2x + 8, x = (3) is a solution because
14 = 2(3) + 8 ?
14 = 6 + 8 ?
14 = 14 yes
Linear equations are the easiest equations to solve and the
simplest of these are the ones where the a single operation is
applied to x. These are the one-step equations. For example,
x – 3 = 12,
12 = x + 3,
3*x = 12,
12 = are one-step equations.x
3
We left the house with a certain amount of money, say $x.
After we bought a hamburger for $4, we had $(x – 4) left.
Suppose there was $14 remained, how much did we have in
the beginning? In symbols, we've the equation x – 4 =14.
Calculate backwards by adding 4 to14, we get 14 + 4 = 18.
In symbols x – 4 =14 add 4 to both sides
+4 +4
so x = 18
We bought 5 apples for $20. How much was each apple?
Again, let x be the price of an apple, then 5 apples cost $5x.
Since the total cost was 20, by dividing 20/5 = 4 and we get
that each apple was $5. In symbols,
5x = 20 both sides divided by 5
5x 20
so x = 4
so = 55
2–2 Linear Equations I

35
Basic principle for solving one- step-equations:
To solve one-step-equations, isolate the x on one side by
applying the opposite operation to both sides of the equation.
Example B. Solve for x
a. x – 3 = 12 Add 3 to both sides
+ 3 + 3
x = 15 check: 15 – 3 = 12
b. x + 3 = –12 Subtract 3 from both sides
–3 –3
x = –15 check: –15 + 3 = –12
3x
3
15
3
=
x = 5 check: 3(5) = 15
c. 3x = 15 Both sides divided by 3
12 = 12 (yes)
?
–12 = –12 (yes)
?
15 = 15 (yes)
?
x
3
–12=d. Multiply both sides by 3
x
3 –12=( (3))
x = –36 Check:
3
–12=– 36
Next we tackle equations that require two steps. These are
the ones that we have to collect the x-terms or the constant
terms first. We do this by addition or subtraction.
Example C. Solve for x
a. x – 6 = 3x Collect the x's by subtracting x from both sides
Divide by 2
Fact: Given a linear equation if we +, –, * , /, to both sides by
the same quantity, the new equation will have the same
solution.
–x –x
–3 = x
–6 2x
=2 2
b. 3x – 6 = 3 collect the numbers by adding 6
+6 +6
3x = 9 div. by 3
3x/3 = 9/3
x = 3
The more general linear equations have the form
#x ±# = #x ±#,
where # can be any number. We solve it by following steps:
1. Add or subtract to move the x-term to one side of the
equation and get: #x ±# = # or # = #x ±#
2. Add or subtract the # to separate the number-term from the
x-term to get: #x = # or # = #x.
3. Divide or multiply to get x:
x = solution or solution = x
The idea is to use + or –, to separate the x-term from the
number term, then use * or / to find the value of x.

36
Exercise
A. Solve in one step by addition or subtraction .
Linear Equations I
1. x + 2 = 3 2. x – 1 = –3 3. –3 = x –5
4. x + 8 = –15 5. x – 2 = –1/2 6. = x –3
2
2
1
B. Solve in one step by multiplication or division.
7. 2x = 3 8. –3x = –1 9. –3 = –5x
10. 8 x = –15 11. –4 =
2
x 12. 7 =
3
–x
13. = –4
3
–x 14. 7 = –x 15. –x = –7
C. Solve by collecting the x‟s to one side first. (Remember to
keep the x‟s positive.)
16. x + 2 = 5 – 2x 17. 2x – 1 = – x –7 18. –x = x – 8
19. –x = 3 – 2x 20. –5x = 6 – 3x 21. –x + 2 = 3 + 2x
22. –3x – 1= 3 – 6x 23. –x + 7 = 3 – 3x 24. –2x + 2 = 9 + x
Example D.
Solve 3x – 4 = 5x + 2
–3x –3x
– 4 = 2x + 2
–2 –2
– 6 = 2x
2
–6
2
2x
=
–3 = x
subtract 3x to remove
the x from one side.
subtract 2 to move the 2 to
the other side.
divide by 2 get x.
Finally, all linear equations can be reduced to the format
#x ±# = #x ±#
by simplifying each side by itself first.
Example E. Solve the equation.
2x – 3(1 – 3x) = 3(x – 6) – 1
Simplify each side by combining like-term first.
2x – 3(1 – 3x) = 3(x – 6) – 1
2x – 3 + 9x = 3x – 18 – 1
11x – 3 = 3x – 19
–3x –3x
8x – 3 = –19
+3 +3
8x = –16
gather the x‟s, subtract 3x from
both sides
gather the number-terms, +3 to
both sides
divide by 8 to get x.
=8 8
8x -16
-2
x = -2

37
D. Solve for x by first simplifying the equations to the form of
#x ± # = #x ± #.
25. 2(x + 2) = 5 – (x – 1) 26. 3(x – 1) + 2 = – 2x – 9
30. –2(x – 3) + 3 = 2(x – 1) + 3x + 13
31. –(x + 4) – 2(x+ 1) = 4(x – 1) – 2
32. x + 1 + 2(x – 3) = 2(x – 1) – (2 – 2x)
33. 4 – 3(2 – 2x) = 2(4x + 1) – 14
34. 5(x – 2) – 3(3 –x) = – 3(x +2) + 2(4x + 1)
35. –3(2 – 2x) + 3(3 – x) = 5(x – 1) + 2(2 – 3x)
27. –2(x – 3) = 2(–x – 1) + 3x
28. –(x + 4) – 2 = 4(x – 1)
29. x + 2(x – 3) = 2(x – 1) – 2
36. 6(2x – 5) – 4 (3x +2) = 2x + 6(–3x – 4) – 8
In this section we make some remarks about solving
equations. Recall that to solve linear equations we simplify
each side of equation first then we separate the x's from the
numbers.
I. (Change-side-change-sign Rule)
When gathering the x's to one side of the equation and the
number terms to the other side, just move terms across to
other side and switch their signs (instead of adding or
subtracting to remove terms).
x + a = b x – a = b
x = b – a x = b + a
2–3 Linear Equations II
change-side
change-sign
change-side
change-signor
There are two sides to group the x's, the right hand side or the
left hand side. It‟s better to collect the x's to the side so that it
ends up with positive x-term.

38
b. 3(x + 2) – x = 2(–2 + x) + x
3x + 6 – x = –4 + 2x + x
2x + 6 = –4 + 3x
4 + 6 = 3x – 2x
10 = x
simplify each side
move 2x and –4 to the other
sides, change their signs
c. + x =
4
3 move to the other side,
change its sign
3
42
1
x = 2
1
4
3
x =
x =
4
–1
–
4
2
4
3–
Example A. Solve.
a. –5 = x + 6
–6 –5 = x
–11 = x
move +6 to the left,
it turns into –6
4
3
4
3
Recall that an equation or a relation expressed using fractions
may always be restated in a way without using fractions.
Example C. The value of 2/3 of an apple is the same as the
value of ¾ of an orange, restate this relation in whole numbers.
2
3 A = 3
4 G,
34
8A = 9G
( ) *12
We've
2
3 A =
3
4 G
or the value of 8 apples equal to 9 oranges.
LCD =12, multiply 12 to both sides,
Rule II. If –x = c, then x = –c
Example B.
a. If –x = –15
then x = 15
b. If –x =
then x = –
Example D. Solve the equations
x4 3
1 2
6
5
4
3 the LCD = 12, multiply it to both
sides to clear the denominators
( ) *12
3 4 2 3
3x + 8 = 10x – 9
9 + 8 = 10x – 3x
17 = 7x
= x
7
17
–+ = x
x4 3
1 2
6
5
4
3–+ = x
b.
x = –64
3
x = –64
3
3x = –24
x = –8
)( 4
a. clear the denominator,
multiply both sides by 4
move 3x to the right and –9
to the left and switch signs.
div by 3
div by 7
III. (Fractional Equations Rule) Multiply fractional equations
by the LCD to remove the fractions first, then solve.

39
(x – 20) = x – 27
100
15
100
45 multiply 100 to both sides
to remove denominators
[ (x – 20) = x – 27
100
15
100
45
] * 100
1 1 100
15 (x – 20) = 45x – 2700
15x – 300 = 45x – 2700
2700 – 300 = 45x –15x
2400 = 30x
2400 / 30 = x
80 = x
c.
(x – 100) = x – 1100
25
100
10 multiply 100 to both sides
to remove denominators
d. 0.25(x – 100) = 0.10x – 1
Change the decimal into fractions, we get
You finish it…
Ans: x =160
14x – 49 = 70x – 98 divide each term by 7
Example. E. Simplify the equation first then solve.
14x – 49 70x – 98=
7 7 7 7
2x – 7 = 10x – 14
14 – 7 = 10x – 2x
7 = 8x
= x8
7
We should always reduce the equation first if it is possible.
IV. (Reduction Rule) Use division to reduce equations to
simpler ones.
Specifically divide the common factor of the coefficients
of each term to make them smaller and easier to work with.

40
There‟re two type of equations that give unusual results.
The first type is referred to as identities. A simple identity is the
equation “x = x”. This corresponds to the trick question “What
number x is equal to itself?” The answer of course is that x
can be any number or that the solutions of equation
“x = x” are all numbers.
Example F. Solve.
2(x – 1) + 3 = x – (– x –1)
This is also the case for the any equation where both sides are
identical such as 2x + 1 = 2x + 1,1 – 4x = 1 – 4x etc…
expand
2x – 2 + 3 = x + x + 1 simplify
2x + 1 = 2x + 1
So this equation is an identity and every number is a solution.
An equation with identical expressions on both sides or can be
rearranged into identical sides has all numbers as its solutions.
Such an equation is called an identity.
two sides are identical
At the opposite end of the identities are the “impossible”
equations where there is no solution at all. An example is the
equation x = x + 1. This corresponds to the trick question
“What number is still the same after we add 1 to it?”
Of course there no such number.
Example F. Solve the equation
2(x – 1) + 4 = x – (– x –1)
If we attempt to solve x = x + 1
x – x = 1
0 = 1we get
expand
2x – 2 + 4 = x + x + 1 simplify
2x + 2 = 2x + 1
2 = 1
So this is an inconsistent equation and there is no solution.
which is an impossibility. These equations are called
inconsistent equations.
Exercise.
A. Solve for x using the switch-side-switch-sign rule.
Remember to move the x‟s first and get positive x‟s.
1. x + 2 = 5 – 2x 2. 2x – 1 = – x – 7 3. –x = x – 8
4. –x = 3 – 2x 5. –5x = 6 – 3x 6. –x + 2 = 3 + 2x
7. –3x – 1= 3 – 6x 8. –x + 7 = 3 – 3x 9. –2x + 2 = 9 + x
Linear Equations II
x = 5 2x = – 7 = x
B. Solve the following fractional equations by using the LCD
to remove the denominators first.
x = –2
5
310. x = –5
3
–411. =
4
3x12.
2
–1
=
3
9x
13. 2
3 x =
3
–2
14. 2
–1
=
6
7x
15. 4
–3

41
x6 3
1 2
3
5
2
3–+ = x16. x4 6
–3 1
8
–5 – 1– = x17.
x4 5
3 2
10
7
4
3+– = x18. x8 12
–5 7
16
–5
+ 1+ = x19.
(x – 20) = x – 3
100
30
100
2020.
(x + 5) – 3 = (x – 5)
100
25
100
2021.
(x +15) = x + 1
100
15
100
3522.
(50 – x) + 2 = (x – 50)
100
25
100
2023.
C. Reduce the equations then solve.
28. –3x – 12 = 30 – 6x 29. 15x – 10x = 25x – 20
30. –4(x – 3) = 12(x + 2) – 8x 31. 15x – 10(x + 2) = 25x – 20
24. –0.3x – 0.25 = 1 – 0.6x 25. 0.15x – 0.1x = 0.25x – 2.4
26. 0.37 – 0.17x = 0.19x – 0.1 27. 1.7x – 0.11 = 0.22 – 0.4x
D. Identify which equations are identities and which are
inconsistent.
34. –x + 1 = 5x – 2(3x + 1) 35. –2(x + 2) = 5x – (7x – 4)
36. 4(x – 3) – 2 = 1 – (14 – 4x) 37. 4(x – 3) – 2 = 1 – (15 – 4x)
32. –x + 1 = –x + 2 33. x + 2 = 5 – (3 – x)
The following are some key words translated into
mathematical operations.
+: add, sum, plus, total, combine, increased by, # more than ..
–: subtract, difference, minus, decreased by, # less than ..
*: multiply, product, times, “fractions or %” of the amount ..
/: divide, quotient, shared equally, ratio ..
Twice = Double = 2* (amount) Square = (amount)2
2–4 Linear Word-Problems
Example A. Let x be an unknown number, write the
mathematical expressions represent the following amounts.
a.300 more than twice x
“2x + 300”
b. 2/3 of x
c. 40% of the sum of x and 1000
3
“ 2 ”x
100
“ 40 ”( x+1000)

42
Follow the following steps to solve word problems.
• Identify the unknown quantities.
• Set x to be the unknown quantity where all the other
unknown quantities in the problem are described in terms
of it. Express them in terms of the chosen x.
• Organize information into a table if possible.
• Translate the problem into an equation.
• Solve the equation.
Example A. Mary and Joe share $10, Mary gets $9 more than
Joe. How much does each get?
Let x = $ Joe gets, then Mary gets (x + 9). They‟ve $10 in total.
x + (x + 9) = 10
2x + 9 = 10
2x = 10 – 9
2x = 1
x = ½ (for Joe) 9 + ½ = 9½ (for Mary)
Example B.
Mary and Joe are to share $120, Joe gets 2/3 of what Mary
gets. How much does each get?
Mary gets $ x, so Joe gets $
3
2 x
3
2 x + x = 120 multiply both sides by 3
3
2 x + x = 120( ) * 3
3 3
2x + 3x = 360
5x = 360
x = 360/5
x = 72
Therefore Mary gets $72 and the rest 120 – 72 = $48 is the
amount for Joe.
Together they have $120, so
Concentration Formula
The concentration of a chemical in a solution is usually
given in %. Let
r = concentration (in %)
S = amount of solution
C = amount of chemical
Then r*S = C
100
25
* 60 = = 15 gallons of salt
3
5 5
25*35
Example C. We have 60 gallons brine (salt water) of 25%
concentration. How much salt is there?
r = concentration = 25% =
S = amount of brine = 60 gal
C = amount of salt = ?
100
25
rS =

43
Example D. (Mixture problem) How many gallons of 10% brine
must be added to 30 gallons of 40% brine to dilute the mixture
to 20% brine?
Make the standard table for mixture problems
concentration # of gallons salt (r*S)
10% x
40% 30
20% (x+30)
10
100 x
40
100 * 30
20
100
(x+30)
Total amount
of salt in the final
mixture
Hence 10
100x + 40
100* 30 = 20
100 (x + 30)
10
100x + 40
100* 30 = 20
100(x + 30)[ ] multiply by 100 to
clear denominator.
* 100
10x + 1200 = 20(x+30)
10x + 1200 = 20x + 600
–600 + 1200 = 20x –10x
600 = 10x  60 = x Ans: 60 gallons
Simple Interest Formula
The interest rate of an investment is usually given in %. Let
r = interest rate (in %)
P = principal
I = amount of interest for one year
then r*P = I.
100
6 * 3000 = 6(30) = $180 = interest.
30
Example E. We have $3000 in a saving account that gives 6%
annual interest. How much interest will be there after one year?
r = interest rate =
P = principal
I = amount of interest
100
6
Hence r*P =
Remark: The general formula for simple interest is P*r*t = I
where t is the number of years. Note the similarity to the
concentration formula.
Example F. (Mixed investments) We have $2000 more in a 6%
account than in a 5% account. In one year, their combined
interest is $560. How much is in each account?
We make a table. Let x = amount in the 5% account.
rate principal interest
6% (x + 2000)
5% x
6
100(x+2000)
5
100 x
Total amount
of interest is
$560
Hence:
6
100(x + 2000) + 5
100x = 560
[ ] * 1006
100(x + 2000) +
5
100x = 560
10011
6(x + 2000) + 5x = 56000
6x + 12000 + 5x = 56000
11x = 56000 – 12000 = 44000
x = 44000/11 = $4000
}
So we've $4000 in the 5% accnt. and $6000 in the 6% accnt.

44
Exercise A. Express each expression in terms of the given x.
Linear Word-Problems
Let x = $ that Joe has.
1. Mary has $10 less than what Joe has, how much does Mary have?
2. Mary has $10 more than twice what Joe has, how much does Mary have?
3. Mary has twice the amount that‟s $10 more than what Joe has, how much does
Mary have?
4. Together Mary and Joe have $100, how much does Mary have?
5. Mary has $100 less than what Joe has, what is twice what Joe and Mary have
together? Simplify your answer.
6. Mary has $100 more than Joe, what is half of what Joe and Mary have together?
Simplify your answer.
B. Peanuts cost $3/lb, cashews cost $8/lb. We have x lbs of peanuts.
Translate each quantity described below in terms of x.
7. What is the cost for the peanuts?
8. We have 4 more lbs of cashews as peanuts, what is the cost for the cashews?
9. Refer to problem 7 and 8, what is the total cost for all the nuts?
Simplify answer.
10. We have twice as many of cashews as peanuts in weight, what is the total cost
of the nuts? Simplify your answer.
11. We have a total of 50 lb of peanuts and cashews, how many lbs of cashews do
we have? What is the total cost for all the nuts? Simplify you answers.
B. Money Sharing Problems. Select the x. Set up an equation then solve.
12. A and B are to share $120. A gets $30 more than B, how much does each get?
13. A and B are to share $120. A gets $16 less than three times of what B gets, how
much does each get?
14. A, B and C are to share $120. A gets $40 more than what B gets, C gets $10
less than what B gets, how much does each get?
15. A, B and C are to share $120. A gets $20 more than what twice of B gets, C
gets $10 more than what A gets, how much does each get?
16. A and B are to share $120. A gets 2/3 of what B gets, how much does each
get?
17. A, B and C are to share $120. A gets 1/3 of what B gets, C gets ¾ of what B
gets, how much does each get?
C. Wet Mixtures Problems (Make a table for each problem 8 – 19)
18. How many gallons of 20% brine must be added to 30 gallons of 40% brine to
dilute the mixture to a 25% brine?
19. How many gallons of pure water must be added to 40 gallons of 30% brine to
dilute the mixture to a 20% brine?
20. How many gallons of 10% brine must be added with 40% brine to make 30
gallons of 20% brine?
21. How many gallons of 50% brine must be added with pure water to make 30
gallons of 20% brine?
22. We have $3000 more saved in an account that gives 5% interest rate than in an
account that gives 4% interest rate. In one year, their combined interest is $600.
How much is in each account?
23. We have saved at a 6% account $2000 less than twice than in a 4% account. In
one year, their combined interest is $1680. How much is in each account?
24. The combined saving in two accounts is $12,000. One account gives 5%
interest rate, the other gives 4% interest rate. The combined interest is %570 in one
year. How much is in each account?
interest rate, the other gives 3% interest rate. The combined interest is $750 in one
year. How much is in each account?
E. Dry Mixture Problem – These are similar to the wet–mixture problems.
26. Peanuts cost $3/lb, cashews cost $6/lb. How many lbs of peanuts are needed
to mix with12 lbs cashews to make a peanut–cashews–mixture that costs $4/lb?
27. Peanuts cost $3/lb, cashews cost $6/lb. How many lbs of each are needed to
make 12 lbs of peanut–cashews–mixture that costs $4/lb?
28. Peanuts cost $2/lb, cashews cost $8/lb. How many lbs of cashews are needed
to mix with 15 lbs of peanuts to get a peanut–cashews–mixture that costs $6/lb?
29. Peanuts cost $2/lb, cashews cost $8/lb. How many lbs of each are needed to
make 50 lbs of peanut–cashews–mixture that cost $3/lb?
D. Mixed Investment Problems.

45
Given an equation with many variables, to solve for a particular
variable means to isolate that variable to one side of the
equation. We do this, just as solving equations in x, by
+, –, * , / the same quantities to both sides of the equations.
These quantities may be numbers or variables.
Example A.
a. Solve for x if x + b = c
Remove b from the LHS by subtracting from both sides
x + b = c
–b –b
x = c – b
b. Solve for w if yw = 5.
Remove y from the LHS by dividing both sides by y.
yw = 5
yw/y = 5/y
w = 5
y
2–5 Literal Equations
To solve for a specific variable in a simple literal equation, do
the following steps.
1. If there are fractions in the equations, multiple by the
LCD to clear the fractions.
2. Isolate the term containing the variable we wanted to
solve for – move all the other terms to other side of the
equation.
3. Isolate the specific variable by dividing the rest of the
factor to the other side.
Example B.
a. Solve for x if (a + b)x = c
(a + b) x = c
x = c
(a + b)
div the RHS by (a + b)
Adding or subtracting a term to both sides may be viewed as
moving the term across the " = " and change its sign.
b. Solve for w if 3y2w = t – 3
3y2w = t – 3
c. Solve for a if b2 – 4ac = 5
b2 – 4ac = 5
– 4ca = 5 – b2
w = t – 3
3y2
a =
5 – b2
–4c
div the RHS by 3y2
move b2 to the RHS
div the RHS by –4c
d. Solve for y if a(x – y) = 10
a(x – y) = 10 expand
ax – ay = 10 subtract ax
– ay = 10 – ax div by –a
y =
10 – ax
–a

46
–4 =
d
3d + b
Multiply by the LCD to get rid of the denominator then solve.
multiply by the LCD d
d
– 4 = d
3d + b
– 4 =
d
3d + b
( )
d
– 4d = 3d + b
– b = 3d + 4d
move –4d and b
– b = 7d
= d
7
–b
Example C.
Solve for d if
div. by 7
Exercise. Solve for the indicated variables.
1. a – b = d – e for b. 2. a – b = d – e for e.
3. 2*b + d = e for b. 4. a*b + d = e for b.
5. (2 + a)*b + d = e for b. 6. 2L + 2W = P for W
7. (3x + 6)y = 5 for y 8. 3x + 6y = 5 for y
w = t – 3
6
13. for t w = t – b
a
14. for t
w =11. for t w =12. for t
t
6
6
t
w = 3t – b
a15. for t 16. (3x + 6)y = 5 for x
w = t – 3
6
17. + a for t w = at – b
5
18. for t
w = at – b
c
19. + d for t 3 = 4t – b
t
20. for t
9. 3x + 6xy = 5 for y 10. 3x – (x + 6)y = 5z for y
We associate each real number with a position on a line,
positive numbers to the right and negative numbers to the left.
-2 20 1 3
+
-1-3
–
2/3 2½ π  3.14..
This line with each position addressed by a real number is
called the real (number) line.
2–6 Inequalities
+–
R
We write this as L < R and called this the natural form because
it corresponds to their respective positions on the real line.
This relation may also be written as R > L (less preferable).
L
<
–π  –3.14..
Given two numbers corresponding to two points on the real
line, we define the number to the right to be greater than the
number to the left.

47
Example A. 2 < 4, –3< –2, 0 > –1 are true statements
and –2 < –5 , 5 < 3 are false statements.
If we want all the numbers greater than 5, we may denote them
as "all number x where 5 < x". In general, we write "a < x" for all
the numbers x greater than a (excluding a). In picture,
+–
a
open dot
If we want all the numbers x greater than or equal to a
(including a), we write it as a < x. In picture
+–
a
solid dot
a < x
a < x
The numbers x fit the description a < x < b where a < b are all
the numbers x between a and b. A line segment as such is
called an interval.
+–
a a < x < b b
Example B.
a. Draw –1 < x < 3.
0 3
+
-1
– x
b. Draw 0 > x > –3
0
+
-3
–
x
Expressions such as 2 < x > 3 or 2 < x < –3 do not have
any solution.
For example 6 < 12, then 6 + 3 < 12 + 3.
We use the this fact to solve inequalities.
Adding or subtracting the same quantity to both retains the
inequality sign, i.e. if a < b, then a ± c < b ± c.
It‟s in the natural form. Mark the numbers and x on the line
in order accordingly.
Put it in the natural form –3 < x < 0.
Then mark the numbers and x in order accordingly.
Example C. Solve x – 3 < 12 and draw the solution.
x – 3 < 12 add 3 to both sides
x – 3 + 3 < 12 + 3
x < 15
0 15
+–
x
A number c is positive means that 0 < c. We may multiply or
divide a positive number to the inequality and keep the same
inequality sign, i.e. if 0 < c and a < b, then ac < bc.
Example D. Solve 3x > 12 and draw the solution.
3x > 12 divide by 3 and keep the inequality sign
3x/3 > 12/3
x > 4 or 4 < x
40
+–
For example 6 < 12 is true, then multiplying by 3
3*6 < 3*12 or 18 < 36 is also true.
x

48
A number c is negative means c < 0. Multiplying or dividing by
an negative number reverses the inequality sign,
i.e. if c < 0 and a < b then
ca > cb .
Example E. Solve –x + 2 < 5 and draw the solution.
–x + 2 < 5 subtract 2 from both sides
–x < 3 multiply by –1 to get x, reverse the inequality
–(–x) > –3
x > –3 or –3 < x
0
+
-3
–
60
+–
12–6–12 <<
For example 6 < 12 is true. If we multiply –1 to both sides then
(–1)6 > (–1)12
– 6 > –12 which is true.
Multiplying by –1 switches the left-right positions of the originals.
To solve inequalities:
1. Simplify both sides of the inequalities
2. Gather the x-terms to one side and the number-terms to the
other sides (use the “change side-change sign” rule).
3. Multiply or divide to get x. If we multiply or divide by
negative numbers to both sides, the inequality sign must be
turned around. This rule can be avoided by keeping the
x-term positive.
Example F. Solve 3x + 5 > x + 9
3x + 5 > x + 9 move the x and 5, change side-change sign
3x – x > 9 – 5
2x > 4 div. 2
20
+–
2x
2
4
2>
x > 2 or 2 < x
Example G. Solve 3(2 – x) > 2(x + 9) – 2x
3(2 – x) > 2(x + 9) – 2x simplify each side
6 – 3x > 2x + 18 – 2x
6 – 3x > 18 move 18 and –3x (change sign)
6 – 18 > 3x
– 12 > 3x
0
+
–12
3
3x
3
>
-4
div. by 3 (no need to switch >)
We also have inequalities in the form of intervals. We solve
them by +, –, * , / to all three parts of the inequalities.
Again, we + or – remove the number term in the middle first,
then divide or multiply to get x. The answer is an interval of
numbers.
–4 > x or x < –4

49
Example H. (Interval Inequality)
Solve 12 > –2x + 6 > –4 and draw.
12 > –2x + 6 > –4 subtract 6
–6 –6 –6
6 > –2x > –10
0
+
-3 < x < 5
5
div. by -2, switch inequality sign
6
-2
-2x
-2<
-10
-2
<
-3
Inequalities
Exercise. A. Draw the following Inequalities. Indicate clearly
whether the end points are included or not.
1. x < 3 2. –5 ≤ x 3. x < –8 4. x ≤ 12
B. Write in the natural form then draw them.
5. x ≥ 3 6. –5 > x 7. x ≥ –8 8. x > 12
C. Draw the following intervals, state so if it is impossible.
9. 6 > x ≥ 3 10. –5 < x ≤ 2 11. 1 > x ≥ –8 12. 4 < x ≤ 2
13. 6 > x ≥ 8 14. –5 > x ≤ 2 15. –7 ≤ x ≤ –3 16. –7 ≤ x ≤ –9
D. Solve the following Inequalities and draw the solution.
17. x + 5 < 3 18. –5 ≤ 2x + 3 19. 3x + 1 < –8
20. 2x + 3 ≤ 12 – x 21. –3x + 5 ≥ 1 – 4x
22. 2(x + 2) ≥ 5 – (x – 1) 23. 3(x – 1) + 2 ≤ – 2x – 9
24. –2(x – 3) > 2(–x – 1) + 3x 25. –(x + 4) – 2 ≤ 4(x – 1)
26. x + 2(x – 3) < 2(x – 1) – 2
27. –2(x – 3) + 3 ≥ 2(x – 1) + 3x + 13
F. Solve the following interval inequalities.
28. –4 ≤
2
x 29. 7 >
3
–x 30. < –4–x
E. Clear the denominator first then solve and draw the solution.
5
x2 3
1 2
3
2
+ ≥ x31. x4
–3
3
–4
– 1> x32.
x2 8
3 3
4
5– ≤33. x8 12
–5 7 1+ >34.
x2 3
–3 2
3
4
4
1–+ x35. x4 6
5 5
3
–1 – 2+ < x36.
x
12 2
7 3
6
1
4
3–– ≥ x37.
≤
40. – 2 < x + 2 < 5 41. –1 ≥ 2x – 3 ≥ –11
42. –5 ≤ 1 – 3x < 10 43. 11 > –1 – 3x > –7
38. –6 ≤ 3x < 12 39. 8 > –2x > –4

50
Chapter 3 – Lines and Linear Equations in x & y
A coordinate system is a system of assigning addresses for
positions in the plane (2 D) or in space (3 D).
The rectangular coordinate system for the plane consists of a
rectangular grid where each point in the plane is addressed by
an ordered pair of numbers (x, y).
3–1 The Rectangular Coordinate System
The horizontal axis is called
the x-axis. The vertical axis
is called the y-axis. The point
where the axes meet
is called the origin.
Starting from the origin, each
point is addressed by its
ordered pair (x, y) where:
x = amount to move
right (+) or left (–).
y = amount to move
up (+) or down (–).
x = amount to move
right (+) or left (–).
y = amount to move
up (+) or down (–).
For example, the point P
corresponds to (4, –3) is
4 right, and 3 down from
the origin.
Example A.
Label the points
A(-1, 2), B(-3, -2),C(0, -5).
The ordered pair (x, y) corresponds to a point is called the
coordinate of the point, x is the x-coordinate and y is the
y-coordinate.
A
B
C
Example B: Find the coordinate of P, Q, R as shown.
P(4, 5), Q(3, -5), R(-6, 0)
P
Q
R
The coordinate of the
origin is (0, 0).
Any point on the x-axis
has coordinate of the
form (x, 0).
Any point on the y-axis
has coordinate of the
form (0, y).
Q1Q2
Q3 Q4
(+,+)(–,+)
(–,–) (+,–)
The axes divide the plane
into four parts. Counter
clockwise, they are denoted
as quadrants I, II, III, and IV.
Respectively, the signs of
the coordinates of each
quadrant are shown.

51
When the x-coordinate of the
a point (x, y) is changed to
its opposite as (–x , y), the
new point is the reflection
across the y-axis.
When the y-coordinate of
the a point (x, y) is changed
to its opposite as (x , –y),
the new point is the
reflection across the x-axis.
(5,4)(–5,4)
(5, –4) (–x, –y) is the reflection of
(x, y) across the origin.
(–5, –4)
Movements and Coordinates
Let A be the point (2, 3).
Suppose it‟s x–coordinate is
increased by 4 to
(2 + 4, 3) = (6, 3) - to the point B,
this corresponds to moving A to the
right by 4.
A B
Similarly if the x–coordinate of
(2, 3) is decreased by 4 to
(2 – 4, 3) = (–2, 3) - to the point C,
this corresponds to moving A to the
left by 4.
Hence we conclude that changes in the x–coordinates of a point
move the point right and left.
If the x–change is +, the point moves to the right.
If the x–change is – , the point moves to the left.
C
x–coord.
increased
by 4
x–coord.
decreased
by 4
(2, 3) (6, 3)(–2, 3)
Again let A be the point (2, 3).
Suppose its y–coordinate is
increased by 4 to
(2, 3 + 4) = (2, 7) - to the point D,
this corresponds to moving A up
by 4.
A
D
Similarly if the y–coordinate of
(2, 3) is decreased by 4 to
(2, 3 – 4) = (2, –1) - to the point E,
this corresponds to
moving A down by 4.
Hence we conclude that changes in the y–coordinates of a point
move the point right and left.
If the y–change is +, the point moves up.
If the y–change is – , the point moves down.
E
y–coord.
increased
by 4
y–coord.
decreased
by 4
(2, 3)
(2, 7)
(2, –1)

52
Example. C.
a. Let A be the point (–2, 4). What is the coordinate of
the point B that is 100 units directly left of A?
Moving left corresponds to decreasing the x-coordinate.
Hence B is (–2 – 100, 4) = (–102, 4).
b. What is the coordinate of the point C that is 100 units
directly above A?
Moving up corresponds to increasing the y-coordinate.
Hence C is (–2, 4) = (–2, 4 +100) = (–2, 104).
c. What is the coordinate of the point D that is 50 to the right
and 30 below A?
We need to add 50 to the x–coordinate (to the right)
and subtract 30 from the y–coordinate (to go down).
Hence D has coordinate (–2 + 50, 4 – 30) = (48, –32).
B. Plot the following points on the graph paper.
2. a. (2, 0) b. (–2, 0) c. (5, 0) d. (–8, 0) e. (–10, 0)
All these points are on which axis?
3. a. (0, 2) b. (0, –2) c. (0, 5) d. (0, –6) e. (0, 7)
All these points are on which axis?
4. a. (5, 2) b. (2, 5) c. (1, 7) d. (7, 1) e. (6, 6)
All these points are in which quadrant?
5. a. (–5, –2) b. (–2, –5) c. (–1, –7) d. (–7, –1) e. (–6, –6)
All these points are in which quadrant?
6. List three coordinates whose locations are in the 2nd
quadrant and plot them.
7. List three coordinates whose locations are in the 4th
quadrant and plot them.
Exercise. A.
a. Write down the coordinates of the following points.
The Rectangular Coordinate System
AB
C
D
E
F
G
H

53
Example A. Graph the points (x, y) where x = –4
(y can be anything).
x y
–4 0
–4 2
–4 4
–4 6
Graph of x = –4
Make a table of
ordered pairs of
points that fit the
description
x = –4.
In the rectangular coordinate system, ordered pairs (x, y)‟s
correspond to locations of points. Collections of points may be
specified by the mathematics relations between the
x-coordinate and the y coordinate. The plot of points that fit a
given relation is called the graph of that relation. To make a
graph of a given mathematics relation, make a table of points
that fit the description and plot them.
3–2 Linear Equations and Lines
x y
-1 -1
0 0
1 1
2 2
Example B. Graph the points (x, y) where y = x.
Make a table of points that fit the description y = x. To find
one such point, we set one of the coordinates to be a
number, any number, than use the relation to find the other
coordinate. Repeat this a few times.
Graph the points (x, y) where y = x
First degree equation in the variables x and y are equations
that may be put into the form Ax + By = C where A, B, C are
numbers. First degree equations are the same as linear
equations. They are called linear because their graphs are
straight lines. To graph a linear equation, find a few ordered
pairs that fit the equation. To find one such ordered pair, assign
a value to x, plug it into the equation and solve for the y (or
assign a value to y and solve for the x). For graphing lines, find
at least two ordered pairs.
Example C.
Graph the following linear equations.
a. y = 2x – 5
Make a table by selecting a few numbers for x. For easy
caluation we set x = -1, 0, 1, and 2. Plug each of these value
into x and find its corresponding y to form an ordered pair.

54
For y = 2x – 5:
x y
-1 -7
0 -5
1 -3
2 -1
If x = -1, then
y = 2(-1) – 5 = -7
If x = 0, then
y = 2(0) – 5 = -5
If x = 1, then
y = 2(1) – 5 = -3
If x = 2, then
y = 2(2) – 5 = -1
b. -3y = 12
Simplify as y = -4
c. 2x = 12
Make a table by
selecting a few
numbers for x.
However, y = -4
is always.
x y
-3 -4
0 -4
3 -4
6 -4
Simplify as x = 6
Make a table.
However the
only selction for
x is x = 6 and y
could be any
number.
x y
6 0
6 2
6 4
6 6
a. y = 2x – 5 b. -3y = 12 c. 2x = 12
If both variables
x and y are
present in the
equation, the
graph is a
tilted line.
If the equation has
only y (no x), the
graph is a
horizontal line.
Summary of the graphs of linear equations:
If the equation has
only x (no y), the
graph is a
vertical line.

55
x y
0 -4
6 0
The x-Intercepts is where the line crosses the x-axis. We set
y = 0 in the equation to find the x-intercept.
The y-Intercepts is where the line crosses the y-axis. We set
x = 0 in the equation to find the y-intercept.
y-int
x-int
Example C. Graph 2x – 3y = 12
by the intercept method.
Since two points determine a line, an easy method to
graph linear equations is the intercept method, i.e. plot the
x-intercept and the y intercept and the graph is the line that
passes through them.
If x = 0, we get
2(0) – 3y = 12
so y = -4
If y = 0, we get
2x – 3(0) = 12
so x = 6
Exercise. A. Solve the indicated variable for each equation with
the given assigned value.
1. x + y = 3 and x = –1, find y.
2. x – y = 3 and y = –1, find x.
3. 2x = 6 and y = –1, find x.
4. –y = 3 and x = 2, find y.
5. 2y = 3 – x and x = –2 , find y.
6. y = –x + 4 and x = –4, find y.
7. 2x – 3y = 1 and y = 3, find x.
8. 2x = 6 – 2y and y = –2, find x.
9. 3y – 2 = 3x and x = 2, find y.
10. 2x + 3y = 3 and x = 0, find y.
11. 2x + 3y = 3 and y = 0, find x.
12. 3x – 4y = 12 and x = 0, find y.
13. 3x – 4y = 12 and y = 0, find x.
14. 6 = 3x – 4y and y = –3, find x.
Linear Equations and Lines
B. a. Complete the tables for each equation with given values.
b. Plot the points from the table. c. Graph the line.
15. x + y = 3 16. 2y = 6
x y
-3
0
3
x y
1
0
–1
17. x = –6
x y
0
–1
– 2
18. y = x – 3
x y
2
1
0
19. 2x – y = 2 20. 3y = 6 + 2x
x y
2
0
–1
x y
1
0
–1
21. y = –6
x y
0
–1
– 2
22. 3y + 4x =12
x y
0
0
1

56
C. Make a table for each equation with at least 3 ordered pairs.
(remember that you get to select one entry in each row as
shown in the tables above) then graph the line.
23. x – y = 3 24. 2x = 6 25. –y – 7= 0
26. 0 = 8 – 2x 27. y = –x + 4 28. 2x – 3 = 6
29. 2x = 6 – 2y 30. 4y – 12 = 3x 31. 2x + 3y = 3
32. –6 = 3x – 2y 33.
35. For problems 29, 30, 31 and 32, use the
intercept-tables as shown to graph the lines.
x y
0
0
intercept-table
36. Why can‟t we use the above intercept method
to graph the lines for problems 25, 26 or 33?
37. By inspection identify which equations give
horizontal lines, which give vertical lines and
which give tilted lines.
3x = 4y 34. 5x + 2y = –10
In general, we need one piece of information to solve for one
unknown quantity. If 2 hamburgers cost 4 dollars and x is the
cost of a hamburger, then 2x = 4 or x = 2 $.
If there are two unknowns, we need two pieces of information
to solve them, three unknowns needs three pieces of
information, etc… These lead to systems of equations.
A system of linear equations is a collection two or more linear
equations in two or more variables.
A solution for the system is a collection of numbers, one for
each variable, that works for all equations in the system.
For example,
2x + y = 7
x + y = 5
is a system. x = 2 and y = 3 is a solution because the ordered
pair fits both equations. This solution is written as (2, 3).
{
3–3 Systems of Linear Equations
Suppose two hamburgers and a salad cost $7, and one
hamburger and one salad cost $5.
Let x = cost of a hamburger, y = cost of a salad.
We can translate these into the system.
2x + y = 7
x + y = 5{
The difference between the order is one hamburger and the
difference between the cost is $2, so the hamburger is $2.
In symbols, subtract the equations, i.e. E1 – E2:
Example A.
substitute 2 for x back into E2,
we get: 2 + y = 5
or y = 3
Hence the solution is (2 , 3) or,
$2 for a hamburger, $3 for a salad.
2x + y = 7
) x + y = 5
x + 0 = 2
so x = 2
E1
E2

57
4x + 3y = 18
3x + 2y = 13
Example B.
{
Subtracting the equations now will not eliminate the x nor y.
We have to adjust the equations before we add or subtract
them to eliminate one of the variable.
Suppose we chose to eliminate the y-terms, we find the LCM
of 3y and 2y first. It's 6y. Then we multiply E1 by 2, E2 by (-3),
and add the results, the y-terms are eliminated.
E1
E2
8x + 6y = 36
-9x – 6y = -39
E1*2:
E2*(-3): + )
-x = -3  x = 3
Suppose four hamburgers and three salads is $18, three
hamburger and two salads is $13. How much does each cost?
Let x = cost of a hamburger, y = cost of a salad.
We translate these into the system:
To find y, set 3 for x in E2:
and get 3(3) + 2y = 13
9 + 2y = 13
2y = 4
y = 2
Hence the solution is (3, 2).
Therefore a hamburger cost $3 and a salad cost $2.
3x + 2y = 13
The above method is called the elimination (addition) method.
We summarize the steps below.
1. Line up the x‟s & y‟s to the left and the numbers to the right.
2. Select the variable x or y to eliminate.
3. Find the LCM of the terms with that variable, and convert
these terms to the LCM (by multiplication).
4. Add or subtract the adjusted equations and solve the
resulting equation.
5. Substitute the answer back into any equation to solve for the
other variable.
5x + 4y = 2
2x – 3y = -13
Solve E1
E2
Eliminate the y. The LCM of the y-terms is 12y.
Multiply E1 by 3 and E2 by 4.
{Example C.
15x + 12y = 6
) 8x – 12y = - 52
3*E1:
4*E2:
Add: 23x + 0 = - 46  x = - 2
Set -2 for x in E1 and get 5(-2) + 4y = 2
-10 + 4y = 2
4y = 12
y = 3
Hence the solution is (-2, 3).
In all the above examples, we obtain exactly one solution in
each case. However, there are two other possibilities.
The system might not have any solution or the system might
have infinite many solutions.

58
Two Special Cases:
I. Contradictory (Inconsistent) Systems
Example D.
{ x + y = 2
x + y = 3
E1 – E2 : x + y = 2
) x + y = 3
0 = -1
(E1)
(E2)
Remove the x-terms by subtracting the equations.
This is impossible! Such systems are said to be contradictory
or inconsistent.
These system don‟t have solution.
Two Special Cases:
I. Contradictory (Inconsistent) Systems
Example D.
{ x + y = 2
x + y = 3
E1 – E2 : x + y = 2
) x + y = 3
0 = -1
(E1)
(E2)
Remove the x-terms by subtracting the equations.
This is impossible! Such systems are said to be contradictory
or inconsistent.
These system don‟t have solution.
Exercise. A. Select the solution of the system.
{x + y = 3
2x + y = 4
Systems of Linear Equations
{x + y = 3
2x + y = 5
{x + y = 3
2x + y = 6
{x + y = 3
2x + y = 31. a. (0, 3) b. (1, 2) c. (2, 1) d.(3, 0)
2. a. (0, 3) b. (1, 2) c. (2, 1) d.(3, 0)
3. a. (0, 3) b. (1, 2) c. (2, 1) d.(3, 0)
4. a. (0, 3) b. (1, 2) c. (2, 1) d.(3, 0)
B. Solve the following systems.
5. {y = 3
2x + y = 3
6. {x = 1
2x + y = 4
7. {y = 1
2x + y = 5
8. {x = 2
2x + y = 6
C. Add or subtract vertically. (These are warm ups)
8x
–9x+)
9. 8
–9–)
10. –8y
–9y–)
11. –8
9–)
12. –4y
–5y+)
13. +4x
–5x–)
14.

59
18.{–x + 2y = –12
2x + y = 4
D. Solve the following system by the elimination method.
15.{x + y = 3
2x + y = 4
16. 17.{x + 2y = 3
2x – y = 6
{x + y = 3
2x – y = 6
19. {3x + 4y = 3
x – 2y = 6
20. { x + 3y = 3
2x – 9y = –4
21.{–3x + 2y = –1
2x + 3y = 5
22. {2x + 3y = –1
3x + 4y = 2
23. {4x – 3y = 3
3x – 2y = –4
24.{5x + 3y = 2
2x + 4y = –2
25. {3x + 4y = –10
–5x + 3y = 7
26. {–4x + 9y = 1
5x – 2y = 8
{
x – y = 3
x – y = –1
3
2
2
3
1
2
1
4
27. {
x + y = 1
x – y = –1
1
2
1
5
3
4
1
6
28.
29.{ x + 3y = 4
2x + 6y = 8
E. Which system is inconsistent and which is dependent?
30.{2x – y = 2
8x – 4y = 6
2x + y = 7 E1
x + y = 5 E2
Solve
by the substitution method.
From E2, x + y = 5, we get x = 5 – y. Then we replace
x by (5 – y) in E1 and get
2(5 – y) + y = 7
10 – 2y + y = 7
10 – y = 7  3 = y
3–4 Systems of Linear Equations II
Example A. {
Put 3 = y back to E2 to find x, we get
x + 3 = 5  x = 2
Therefore the solution is (2, 3)
There are two other methods to solve system of equations.
Substitution Method
In substitution method, we solve for one of the variables in
terms of the other, then substitute the result into the other
equation.
We use the substitution method when it's easy to solve for
one of the variable in terms of the other. Specifically, it is
easy to solve for a variable when an equation contains an
single x or single y.
2x – y = 7 E1
3x + 2y = 7 E2
Solve
by the substitution method.
By inspection, we see that it's easy to solve for the y using E1.
From E1, 2x – y = 7, we get 2x – 7= y. Substitute y by (2x – 7)
in E2 and get
3x + 2(2x – 7) = 7
3x + 4x – 14 = 7
7x = 21
x = 3
Example B. {
To find y, use the substitution equation, set x = 3 in y = 2x – 7
y = 2(3) – 7
y = -1 Hence the solution is (3, -1).

60
Graphing Method
Given a system of linear equations the graph of each equation
is a straight line. The solution of the system is the intersection
point (x, y) of the these lines. Thus we may find the solution by
graphing the lines and locate the point of intersection
graphically. In general, we don't not use the graphing method
because it is not easy to do it accurately.
2x + y = 7
x + y = 5
Solve graphically.{ E1
E2
Example C.
Use intercept method.
x y
0 7
7/2 0
2x + y = 7
(0, 7)
(7/2, 0)
(0, 5)
(5, 0)
The intersection
(2, 3) is the solution
E1
E2
x + y = 5
x y
0 5
5 0
E1(Check another point.)
Graphing of inconsistent systems are parallel lines which do
not intersect, hence there is no solution.
x + y = 7
x y
0 7
7 0
x y
0 5
5 0
(0, 7)
(7, 0)
(0, 5)
(5, 0)
No intersection.
No solution
x + y = 7
x + y = 5
Solve graphically.{ E1
E2
Example D.
x + y = 5
Graphs of dependent systems are identical lines, hence every
point on the line is a solution.
2x + 2y = 10 is the same as
x + y = 5
x y
0 5
5 0
(5, 0)
(0, 5)
Every point is a solution,
e.g.(0, 5), (2, 3), (5, 0)…
(2, 3)
x + y = 5
2x + 2y = 10
Solve graphically.{
E1
E2
Example E.

61
Systems of Linear Equations II
4. {–x + 2y = –12
y = 4 – 2x
Exercise. Solve by the substitution method.
1. {y = 3 – x
2x + y = 4
2. 3. {x = 3 – y
2x – y = 6
{x + y = 3
2x + 6 = y
5. {3x + 4y = 3
x = 6 + 2y
6. { x = 3 – 3y
2x – 9y = –4
10. Graph the inconsistent system
{ x + 3y = 4
2x + 6y = 8
{2x – y = 2
8x – 4y = 6
Problem 7, 8, 9: Solve problem 1, 2, and 3 by graphing.
11. Graph the dependent system
Following are key-words translated into mathematic operations.
+: add, sum, plus, total, combine, increased by, # more than ..
–: subtract, difference, minus, decreased by, # less than ..
*: multiply, product, times, “fractions or %” of the amount ..
/: divide, quotient, shared equally, ratio ..
Twice = Double = 2* (amount)
Square = (amount)2
3–5 Linear Word-Problems II
In chapter 2, we solved problems with a single variable. Many
of those problems may be solved using two variables and a
system of equations. The advantage of using two variables
instead of one is that it's easier to construct the equations to
form the system. To do this:
I. identify the unknown quantities and named them as x and y
II. usually there are two numerical relations given, translate
each numerical relation into an equation to make a system
III. solve the system.
Example A. A 30-foot rope is cut into two pieces. The longer
one is 6 feet less than twice of the short piece. How long is
each piece?
Let x = length of the long piece.
y = length of the short piece
Then x + y = 30 --- E1
x = 2y – 6 --- E2{
Use the substitution method.
Substitute x = (2y – 6) into E1, we get
(2y – 6) + y = 30 solve for y
2y – 6 + y = 30
3y – 6 = 30
3y = 36
y = 36/3 = 12
Put y = 12 into E2, we get x = 2(12) – 6 =18.
Hence the two pieces are 18ft and 12ft.

62
Example B. Suppose four hamburgers and three fries costs
$18, and three hamburgers and two fries costs $13. Find the
price of each.
4x + 3y = 18
3x + 2y = 13
The system is
E1
E2
Use the elimination method. LCM of 3y and 2y is 6y.
Multiply E1 by 2 and E2 by 3, subtract
Let x = price of a hamburger, y = price of fries
8x + 6y = 36
) 9x + 6y = 39
2*E1:
3*E2:
{
–x = –3  x = 3
To find y, put x = 3 into E2, 3(3) + 2y = 13
9 + 2y = 13
2y = 13 – 9
2y = 4 y = 4/2 = 2.
So a hamburger is $3, and fries are $2.
The RTD Formula
Let R = Rate or speed (usually mph)
T = Time (in hours)
D = Distance (in miles)
then RT = D.
For example, if we drove from LA to SF at a rate of 75 mph
and it took 6 hours, it means R = 75, T = 6. Therefore
75(6) = 450 miles is the distance between the two cities.
Following problems use formulas in setting up the equations.
Up-and-down-stream-problems
When a boat travels up and down a stream, the speed of the
boat depends on the boat's speed R in still water (cruising
speed) and the current's speed C. Going upstream, the
current slows the boat down and going downstream it speeds
up the boat. In fact, the upstream speed is (R – C), and the
downstream speed is (R + C).
Example C. A boat can travel upstream for 24 miles in 6
hours and the same distance downstream in 3 hours.
What is the cruising speed of the boat and the current speed?
Let R = cruising speed, C = current speed,
Make a table.
Rate Time (T) Distance (D)
Upstream R – C 6 24
Downstream R + C 3 24
By the formula R*T = D we get two equations.
For example, a boat that travels 3 mph in still water goes
upstream against a 1 mph current, it's upstream rate is 2mph.
It would take 6 hours to go 12 miles. If it travels downstream,
it can travel at 4 mph and it would take 3 hours to go 12 miles.

63
6(R – C) = 24
3(R + C) = 24{
Simplify each equation, divide E1 by 6 and divide E2 by 3.
Add the equations to eliminate C
+) 2R = 12
R = 6
Substitute R = 6 into E2,
6 + C = 8
C = 2
R – C = 4 --- E3
R + C = 8 --- E4
R = cruising speed = 6 mph, C = current speed = 2 mph
--- E1
--- E2
{
So,
Simple Interest Formula
Recall the simple interest of an investment is I = R*P where
I = amount of interest for one year
R = interest (in %)
P = principal
Example D. (Mixed investments)
We have $20,000 saved in two accounts which give 5% and
6% interest respectively. In one year, their combined
interest is $1150. How much is in each account?
rate principal interest
6% x
5% y
total 20,000 1150
6
100
5
100 * y
* x
x + y = 20,000 --- E1
6
100
5
100
y = 1150 --- E2x +{
6
100
5
100
y = 1150)100x +(
100
6x + 5y = 115000 ---E3
Make a table using the interest formula R*P = I
Multiply E2 by 100 to clear the denominator.
x + y = 20,000 ---E1
6x + 5y = 115000 ---E3
To eliminate y, multiply E1 by -5 and add to E3.
6x + 5y = 115000
-5x – 5y = -100,000
+) x = 15000
Sub x=15000 into E1,
15000 + y = 20000
y = 5000
So there are $15,000 in the 6% account,
and $5,000 in the 5% account.
Linear Word-Problems II
{
E1(-5):

64
Exercise. A. The problems below are from the word–problems for linear
equations with one variable.
1. A and B are to share $120. A gets $30 more than B, how much does each get?
(Just let A = $ that Mr. A has, and B = $ that Mr. B has.)
2. A and B are to share $120. A gets $30 more than twice of what B gets, how
much does each get?
3. A and B are to share $120. A gets $16 less than three times of what B gets,
how much does each get?
Let x = number of lb of peanuts y = number of lb of cashews
Make tables for problems 4-8. .
4. Peanuts costs $2/lb, cashews costs $8/lb. How many lbs of each are needed
to get 50 lbs of peanut–cashews–mixture that‟s $3/lb?
5. Peanuts costs $3/lb, cashews costs $6/lb. How many lbs of each are needed
to get 12 lbs of peanut–cashews–mixture that„s $4/lb?
6. We have $3000 more saved at an account that gives 5% interest than at an
account that gives 4% interest. In one year, their combined interest is $600. How
much is in each account?
7. We have saved at a 6% account $2000 less than twice than at a 4% account.
In one year, their combined interest is $520. How much is in each account?
interest, the other gives 4% interest. The combined interest is %570 in one year.
How much is in each account?
Linear Word-Problems II
B. Find x = $ of cashews and y = $ of peanuts per lb, solve for x and y with the
following information.
9. Order 1. (in lb) Order 2 (in lb)
Cashews 2 3
Peanuts 3 1
Cost $16 $17
10.
Order 1. (in lb) Order 2 (in lb)
Cashews 2 3
Peanuts 3 1
Cost $17 $15
11. Order 1 consists of 5 lb of cashews and 3 lb of peanuts and it costs $16,
Order 2 consists of 7 lb of cashews and 2 lb of peanuts and it costs $18.
Organize this information into a table and solve foe x and y.
12. Order 1 consists of 3 lb of cashews and 4 lb of peanuts and it costs $38,
Order 2 consists of 2 lb of cashews and 5 lb of peanuts and it costs $30.
Organize this information into a table and solve for x and y.
Note that 9–12 are the same as the hamburger–fries problem.
13. We have $3000 more saved in an account that gives 5% interest than in an
account that gives 4% interest. In one year, their combined interest is $600. How
much is in each account?
14. We have saved in a 6% account $2000 less than twice than in a 4% account. In
one year, their combined interest is $1680. How much is in each account?
15. The combined saving in two accounts is $12,000. One account has 5% interest,
the other has 4% interest. The combined interest is $570 in one year. How much is
in each account?
16. The combined saving in two accounts is $15,000. One account has 6% interest,
the other has 3% interest. The combined interest is $750 in one year. How much is
in each account?
C. Select the x and y, make a table, then solve.

65
Chapter 4 – Polynomial Expressions and Operations
Example A.
43 = (4)(4)(4) = 64
(xy)2= (xy)(xy) = x2y2
xy2 = (x)(yy)
–x2 = –(xx)
base
exponent
4–1 Exponents
Multiplication Rule: ANAK =AN+K
Example B.
a. 5354 = (5*5*5)(5*5*5*5) = 53+4 = 57
b. x5y7x4y6 = x5x4y7y6 = x9y13
Rules of Exponents
Division Rule:
Example C.
AN
AK = AN – K
56
52 =
(5)(5)(5)(5)(5)(5)
(5)(5)
= 56 – 2 = 54
We write the quantity A multiplied to itself N times as AN, i.e.
A x A x A ….x A = AN
Power Rule: (AN)K = ANK
Example D. (34)5 = (34)(34)(34)(34)(34) = 34+4+4+4+4 = 34*5 = 320
Since = 1 = A1 – 1 = A0, we obtain the 0-power Rule.
A1
A1
Since = = A0 – K = A–K, we get the negative-power Rule.
1
AK
A0
AK
Negative-Power Rule: A–K = 1
AK
Example D. Simplify
1
32
1
9
c. ( )–12
5 = 1
2/5 = 1*
5
2 =
5
2
b. 3–2 = =
a. 30 = 1
In general ( )–Ka
b = ( )K
b
a
d. ( )–22
5
= ( )2 = 25
4
5
2
0-Power Rule: A0 = 1, A = 0
, A = 0
e. 3–1 – 40 * 2–2 =
Although the negative power means to reciprocate,
for problems of consolidating exponents, we do not
reciprocate the negative exponents. Instead we add or
subtract them using the multiplication and division rules first.
= x4 – 8 y–6+23
= x–4 y17
= y17
=
Example E. Simplify 3–2 x4 y–6 x–8 y 23
3–2 x4 y–6 x–8 y23
= 3–2 x4 x–8 y–6 y23
1
9
1
9
1
9x4
y17
9x4
1
3 – 1*
1
22 = 1
3
– 1
4
= 1
12

66
Example F. Simplify using the rules for exponents.
Leave the answer in positive exponents only.
23x–8
26 x–3
23x–8
26x–3
= 23 – 6 x–8 – (–3 )
= 2–3 x–5
=
23
1
x5
1
* = 8x5
1
Example G. Simplify (3x–2y3)–2 x2
3–5x–3(y–1x2)3
= 3–2x4y–6x2
3–5x–3y–3 x6 =
= = 3–2 – (–5) x6 – 3 y–6 – (–3)
= 33 x3 y–3 = 27 x3
(3x–2y3)–2 x2
3–5x–3(y–1x2)3 3–5x–3x6y–3
3–2x4x2y–6
3–2x6y–6
3–5x3y–3
y3
Exercise. A. Write the numbers without the negative
exponents and compute the answers.
1. 2–1 2. –2–2 3. 2–3 4. (–3)–2 5. 3–3
6. 5–2 7. 4–3 8. 1
2
( )
–3
9. 2
3
( )
–1
10. 3
2
( )
–2
11. 2–1* 3–2 12. 2–2+ 3–1 13. 2* 4–1– 50 * 3–1
14. 32 * 6–1– 6 * 2–3 15. 2–2* 3–1 + 80 * 2–1
B. Combine the exponents. Leave the answers in positive
exponents–but do not reciprocate the negative exponents until
the final step.
16. x3x5 17. x–3x5 18. x3x–5 19. x–3x–5
20. x4y2x3y–4 21. y–3x–2 y–4x4 22. 22x–3xy2x32–5
23. 32y–152–2x5y2x–9 24. 42x252–3y–34 x–41y–11
25. x2(x3)5 26. (x–3)–5x –6 27. x4(x3y–5) –3y–8
Exponents
x–8
x–3
B. Combine the exponents. Leave the answers in positive
exponents–but do not reciprocate the negative exponents until
the final step.
28. x8
x–329.
x–8
x330. y6x–8
x–2y331.
x6x–2y–8
y–3x–5y232.
2–3x6y–8
2–5y–5x233.
3–2y2x4
2–3x3y–234.
4–1(x3y–2)–2
2–3(y–5x2)–135.
6–2 y2(x4y–3)–1
9–1(x3y–2)–4y236.
C. Combine the exponents as far as possible.
38. 232x 39. 3x+23x 40. ax–3ax+5
41. (b2)x+1b–x+3 42. e3e2x+1e–x
43. e3e2x+1e–x
44. How would you make sense of 23 ?
2

67
100 = 1
101 = 10
102 = 100
103 = 1000
10–1 = 0.1
10–2 = 0.01
10–3 = 0.001
10–4 = 0.0001
4–2 Scientific Notation
Scientific Notation
Any number can be written in the form
A x 10N
where 1 < A < 10. This form is called the scientific notation of
the number.
An important application for exponents is the use of the
powers of 10 in calculation of very large or very small numbers.
Powers of 10:
To write a number in scientific notation, we move the decimal
point behind the first nonzero digit.
i. If the decimal point moves to the left N spaces, then the
exponent over 10 is positive N, i.e. if after moving the
decimal point we get a smaller number A, then N is positive.
ii. If the decimal point moves to the right N spaces, then the
exponent over 10 is negative, i.e. if after moving the decimal
point we get a larger number A, then N is negative.
Move left 4 places.
Move right 3 places
Example A. Write the following numbers in scientific
notation.
a. 12300. = 1 2300 . = 1. 23 x 10 +4
b. 0.00123 = 0. 001 23 = 1. 23 x 10 –3
To change a number in scientific notation back to the standard
form, we move the decimal point according to N.
i. If N is positive, move the decimal point in A to the right,
i.e. make A into a larger number.
ii. If N is negative, move the decimal point in A to the left,
i.e. make A into a smaller number.
Move right 4 places,
Move left 3 places
Example B. Write the following numbers in the standard form.
a. 1. 23 x 10 +4 = 1 2300 . = 12300.
b. 1. 23 x 10 –3 = 0. 001 23 = 0.00123
Scientific notation simplifies multiplication and division of
very large and very small numbers.

68
Example C. Calculate. Give the answer in both scientific
notation and the standard notation.
a. (1.2 x 108) x (1.3 x 10–12)
= 1.2 x 1.3 x 108 x 10 –12
= 1.56 x 108 –12
= 1.56 x 10 –4
= 0.000156
b. 6.3 x 10-2
2.1 x 10-10
=
6.3
2.1
x 10 – 2 – ( – 10)
= 3 x 108
= 300,000,000
Example D. Convert each numbers into scientific notation.
Calculate the result. Give the answer in both scientific
notation and the standard notation.
240,000,000 x 0.0000025
=
0.00015
2.4 x 108 x 2.5 x 10–6
1.5 x 10–4
= 2.4 x 2.5
1.5
x 10 8 + (–6) – ( – 4)
= 2.4 x 2.5 x 108 x 10–6
1.5 x 10–4
= 4 x 108 – 6 + 4
= 4 x 106 = 4,000,000

69
Ex.ercise A. Convert the following numbers into standard
notation. Remember that multiplying by 10positive power makes a
number larger, and multiplying by 10negative power makes the
number smaller.
Scientific Notation
1. 2.41 x 103 2. 2.41 x 10–3 3. 315 x 10–2 4. 315 x 102
5. 0.762 x 10–5 7. 5.93 x 10–66. 0.762 x 105
8. 593000 x 10–6
B. Convert the following numbers in scientific notation into
standard notation.
9. 5.31 x 103 10. 2.41 x 10–3 11. 3.1 x 10–2
12. 9 x 102 13. 5.89 x 109 14. 7.11 x 10–6
15. 5.2 x 10–9 16. 7.11 x 1011
C. Convert the following numbers into the scientific notation.
17. 10 18. 20 19. 200
20. 2,310 21. 231,000 22. 231,000,000,000,000
23. 1 24. 0.1 25. 0.02
27. 0.00000231 28. 0.000000000000231
26. 0.0002
C. Convert the following numbers into the scientific notation.
29. 0.023128. 23,100 30. 2.31
D. Calculate by converting into the scientific notation first. Put
the answers in both the scientific and standard notations.
31. 100 x 0.01 32. 100 x 0.1 33. 100 x 0.001
34. 4,000 x 1,200,000 35. 0.000004 x 0.000012
36. 400,000 x 0.0000012 37. 0.004 x 1,200,000
38. 0.006 x 2,400,000 39. 0.00024 x 600,000
40. 0.0024 x 0.00006 41. 240,000,000 x 600,000,000
240 x 0.000004
0.00016
42.
0.00015
43.
x 0.0008
0.00024 x 600,000
1,400,000,000 x 6,000
0.0000035
44.
150
45.
x 800
0.00024 x 0. 00005
46. Google “scientific notation applications” and list one that
impresses you.

70
Example A.
2 + 3x  “the sum of 2 and 3 times x”
4x2 – 5x  “the difference between 4 times the square of x
and 5 times x”
(3 – 2x)2  “the square of the difference of 3 and twice x”
A mathematics expression is a calculation procedure written in
numbers, variables, and operation symbols.
Example B. Evaluate the monomials if y = –4
a. 3y2
3y2  3(–4)2
= 3(16) = 48
An expression of the form #xN, where the exponent N is a
non-negative integer and # is a number, is called a monomial
(one-term).
For example, 3x2, –4x3, and 5x6 are monomials.
4–3 Polynomial Expressions
b. –3y2 (y = –4)
–3y2  –3(–4)2
= –3(16) = –48.
c. –3y3
–3y3  – 3(–4)3
= – 3(–64) = 192
The sum of monomials are called polynomials (many-terms),
these are expressions of the form
#xN ±#xN-1 ±… ±#x1 ±#
where # can be any number.
For example, 4x + 7, –3x2 – 4x + 7, –5x4 + 1 are polynomials,
x
1 is not a polynomial.whereas the expression
Polynomial Expressions
Example C. Evaluate the polynomial 4x2 – 3x3 if x = –3.
The polynomial 4x2 – 3x3 is the combination of two
monomials; 4x2 and –3x3. When evaluating the polynomial,
we evaluate each monomial then combine the results.
Set x = (–3) in the expression, we get
4(–3)2 – 3(–3)3
= 4(9) – 3(–27)
= 36 + 81
= 117
Given a polynomial, each monomial is called a term.
#xN ±#xN-1 ±… ±#x ±#
terms
Therefore the polynomial –3x2 – 4x + 7 has 3 terms,
–3x2 , –4x and + 7.

71
Each term is addressed by the variable part. Hence the
x2-term of the –3x2 – 4x + 7 is –3x2, the x-term is –4x,
and the number term or the constant term is 7.
The number in front of a term is called the coefficient of that
term. So the coefficient of –3x2 is –3 .
Terms with the same variable part are called like-terms.
Like-terms may be combined.
For example, 4x + 5x = 9x and 3x2 – 5x2 = –2x2.
Unlike terms may not be combined. So x + x2 stays as x + x2.
Note that we write 1xN as xN , –1xN as –xN.
When multiplying a number with a term, we multiply it with the
coefficient. Hence, 3(5x) = (3*5)x =15x,
and –2(–4x) = (–2)(–4)x = 8x.
Operations with Polynomials
When multiplying a number with a polynomial, we may
expand using the distributive law: A(B ±C) = AB ±AC.
Example D. Expand and simplify.
a. 3(2x – 4) + 2(4 – 5x)
= 6x – 12 + 8 – 10x
= –4x – 4
b. –3(x2 – 3x + 5) – 2(–x2 – 4x – 6)
= –3x2 + 9x – 15 + 2x2 + 8x +12
= –x2 + 17x – 3
Polynomials in Two or More Variables
We form polynomials in two variables say, x & y, by adding
monomials of the form kx#y# where k is a number and the
powers are all nonnegative integers such as –5x3y2 or 3x2.
Like–terms are terms where the variable parts are the same.
For example 3x2y3 + 5x2y3 = 8x2y3 but 3x2y3 + 5x3y3 can‟t be
combined. We evaluate them by assigning numbers to
x and/or y. If only one number is given, the result is a formula.
If both numbers are given, then we get a numerical output.
We may do this for x, y and z or even more variables.
Example E. Expand and simplify.
a. 3(2xy – 4x2y) + 2xy – 3xy2
= 6xy – 8x2y + 2xy – 3xy2
= 8xy – 8x2y – 3xy2
b. Evaluate 8xy – 8x2y – 3xy2 if x = 2.
Input x = 2, we get 8(2)y – 8(2)2y – 3(2)y2
= 16y – 32y – 6y2
= –16y – 6y2
c. Evaluate 8xy – 8x2y – 3xy2 if x = 2 and y = 3
We may put x = 2, y = 3 into the formula and do everything
all over again or we may plug into y = 3 into part b which is
easier. We will do the easy way.
–16(3) – 6(3)2
Input y = 3 into –16y – 6y2
we get
= –48 – 54 = –102

72
Exercise. A. Evaluate each monomials with the given values.
3. 2x2 with x = 1 and x = –1 4. –2x2 with x = 1 and x = –1
5. 5y3 with y = 2 and y = –2 6. –5y3 with y = 2 and y = –2
1. 2x with x = 1 and x = –1 2. –2x with x = 1 and x = –1
7. 5z4 with z = 2 and z = –2 8. –5y4 with z = 2 and z = –2
B. Evaluate each monomials with the given values.
9. 2x2 – 3x + 2 with x = 1 and x = –1
10. –2x2 + 4x – 1 with x = 2 and x = –2
11. 3x2 – x – 2 with x = 3 and x = –3
12. –3x2 – x + 2 with x = 3 and x = –3
13. –2x3 – x2 + 4 with x = 2 and x = –2
14. –2x3 – 5x2 – 5 with x = 3 and x = –3
C. Expand and simplify.
15. 5(2x – 4) + 3(4 – 5x) 16. 5(2x – 4) – 3(4 – 5x)
17. –2(3x – 8) + 3(4 – 9x) 18. –2(3x – 8) – 3(4 – 9x)
19. 7(–2x – 7) – 3(4 – 3x) 20. –5(–2 – 8x) + 7(–2 – 11x)
Polynomial Expressions
21. x2 – 3x + 5 + 2(–x2 – 4x – 6)
22. x2 – 3x + 5 – 2(–x2 – 4x – 6)
23. 2(x2 – 3x + 5) + 5(–x2 – 4x – 6)
24. 2(x2 – 3x + 5) – 5(–x2 – 4x – 6)
25. –2(3x2 – 2x + 5) + 5(–4x2 – 4x – 3)
26. –2(3x2 – 2x + 5) – 5(–4x2 – 4x – 3)
27. 4(3x3 – 5x2) – 9(6x2 – 7x) – 5(– 8x – 2)
29. Simplify 2(3xy – xy2) – 2(2xy – xy2) then evaluated it
with x = –1, afterwards evaluate it at (–1, 2) for (x, y)
30. Simplify x2 – 2(3xy – x2) – 2(y2 – xy) then evaluated it
with y = –2, afterwards evaluate it at (–1, –2) for (x, y)
31. Simplify x2 – 2(3xy – z2) – 2(z2 – x2) then evaluated it
with x = –1, y = – 2 and z = 3.
28. –6(7x2 + 5x – 9) – 7(–3x2 – 2x – 7)

73
Terms with the same variabe part are called like-terms.
Like-terms may be combined.
For example, 4x + 5x – 3x2 – 5x2 = 9x – 2x2.
Unlike terms may not be combined.
In particular x2 + x2 = 2x2 = x4, and that x2 + x3 = x2 + x3 = x5.
Note that we write 1xN as xN , –1xN as –xN.
When multiplying a number with a term, we multiply it to the
coefficient. Hence, 3(5x) = (3*5)x =15x,
and –2(–4x) = (–2)(–4)x = 8x.
When multiplying a number to a polynomial, we may expand
the result using the distributive law: A(B ±C) = AB ±AC.
4–4 Polynomial Operations
In the last section, we introduced polynomial expressions.
Given a polynomial, each monomial is also called a term.
#xN ±#xN-1 ±… ±#x ±#
terms
Example A. Expand and simplify.
a. 3(2x – 4) + 2(4 – 5x)
= 6x – 12 + 8 – 10x
= –4x – 4
b. –3(x2 – 3x + 5) – 2(–x2 – 4x – 6)
= –3x2 + 9x – 15 + 2x2 + 8x +12
= –x2 + 17x – 3
When multiply a term with another term, we multiply the
coefficient with the coefficient and the variable with the
variable.
Example B.
a. (3x2)(2x3) = 3*2x2x3 = 6x5
b. 3x2(–4x) = 3(–4)x2x = –12x3
c. 3x2(2x3 – 4x) distribute
= 6x5 – 12x3
To multiply two polynomials, we may multiply each term of one
polynomial against other polynomial then expand and simplify.
Example C.
b. (2x – 1)(2x2 + 3x –4)
= 3x(2x – 1) + 2(2x – 1)
= 6x2 – 3x + 4x – 2
= 6x2 + x – 2
= 2x(2x2 + 3x –4) –1(2x2 + 3x – 4)
= 4x3 + 6x2 – 8x – 2x2 – 3x + 4
= 4x3 + 4x2 – 11x + 4
a. (3x + 2)(2x – 1)
Note that if we did (2x – 1)(3x + 2) or (2x2 + 3x –4)(2x – 1)
instead, we get the same answers. (Check this.)
Fact. If P and Q are two polynomials then PQ ≡ QP.
A shorter way to multiply is to bypass the 2nd step and use the
general distributive law.

74
General Distributive Rule:
(A ±B ±C ±..)(a ±b ±c ..)
= Aa ±Ab ±Ac ..±Ba ±Bb ±Bc ..±Ca ±Cb ±Cc ..
Example D. Expand
a. (x + 3)(x – 4)
= x2 – 4x + 3x – 12 simplify
= x2 – x – 12
b. (x – 3)(x2 – 2x – 2)
= x3 – 2x2 – 2x – 3x2 + 6x + 6
= x3– 5x2 + 4x + 6
We will address the division operation of polynomials later-
after we understand more about the multiplication operation.
Exercise. A. Multiply the following monomials.
1. 3x2(–3x2)
11. 4x(3x – 5) – 9(6x – 7)
Polynomial Operations
2. –3x2(8x5) 3. –5x2(–3x3)
4. –12( )6
–5x3
5. 24( x3)8
–5 6. 6x2( )3
2x3
7. –15x4( x5)5
–2
B. Fill in the degrees of the products.
8. #x(#x2 + # x + #) = #x? + #x? + #x?
9. #x2(#x4 + # x3 + #x2) = #x? + #x? + #x?
10. #x4(#x3 + # x2 + #x + #) = #x? + #x? + #x? + #x?
12. –x(2x + 7) + 3(4x – 2)
13. –3x(3x + 2) – 8x(7x – 5) 14. 5x(–5x + 9) + 6x(6x – 1)
15. 2x(–4x + 2) – 3x(2x – 1) – 3(4x – 2)
16. –4x(–7x + 9) – 2x(2x – 5) + 9(4x + 2)
18. (x + 5)(x + 7)
D. Expand and simplify. (Use any method.)
19. (x – 5)(x + 7)
20. (x + 5)(x – 7) 21. (x – 5)(x – 7)
22. (3x – 5)(2x + 4) 23. (–x + 5)(3x + 8)
24. (2x – 5)(2x + 5) 25. (3x + 7)(3x – 7)
26. (3x2 – 5)(x – 6) 27. (8x – 2)(–4x2 – 7)
28. (2x – 7)(x2 – 3x + 9) 29. (5x + 3)(2x2 – x + 5)
38. (x – 1)(x + 1) 39. (x – 1)(x2 + x + 1)
40. (x – 1)(x3 + x2 + x + 1)
41. (x – 1)(x4 + x3 + x2 + x + 1)
42. What do you think the answer is for
(x – 1)(x50 + x49 + …+ x2 + x + 1)?
30. (x – 1)(x – 1) 31. (x + 1)2
32. (2x – 3)2 33. (5x + 4)2
34. 2x(2x – 1)(3x + 2) 35. 4x(3x – 2)(2x + 3)
36. (x – 5)(2x – 1)(3x + 2) 37. (2x + 1)(3x + 1)(x – 2)

75
A binomial is a two-term polynomial. Usually we use the
term for expressions of the form ax + b.
A trinomial is a three term polynomial. Usually we use the
term for expressions of the form ax2 + bx + c.
The product of two binomials is a trinomial.
(#x + #)(#x + #) = #x2 + #x + #
4–5 Special Binomial Operations
F: To get the x2-term, multiply the two Front x-terms of the
binomials.
OI: To get the x-term, multiply the Outer and Inner pairs and
combine the results.
L: To get the constant term, multiply the two Last constant
terms.
This is called the FOIL method.
The FOIL method speeds up the multiplication of above
binomial products and this will come in handy later.
b. (3x + 4)(–2x + 5) = –6x2 + 7x + 20
Example A. Multiply using FOIL method.
a. (x + 3)(x – 4) = x2 – x – 12
The last terms: 20
The last terms: –12
Expanding the negative of the binomial product requires
extra care. One way to do this is to insert a set of “[ ]”
around the product.
Example B. Expand.
a. – [(3x – 4)(x + 5)]
= – [ 3x2 + 15x – 4x – 20]
= – [ 3x2 + 11x – 20]
= – 3x2 – 11x + 20
Insert [ ]
Expand
Remove [ ] and
change signs.
Another way to do this is to distribute the negative sign into
the first binomial then FOIL.
Example C. Expand.
a. – (3x – 4)(x + 5)
= (–3x + 4)(x + 5)
= – 3x2 – 15x + 4x + 20
= – 3x2 – 11x + 20
Distribute the sign.
Expand
b. Expand and simplify. (Two versions)
(2x – 5)(x +3) – (3x – 4)(x + 5)
= (2x – 5)(x +3) + (–3x + 4)(x + 5)
= 2x2 + 6x – 5x – 15 + (–3x2 –15x + 4x + 20)
= 2x2 + x – 15 – 3x2 – 11x + 20
= –x2 – 10x + 5
(2x – 5)(x +3) – [(3x – 4)(x + 5)] Insert brackets
= 2x2 + x – 15 – [3x2 +11x – 20]
= 2x2 + x – 15 – 3x2 – 11x + 20
= –x2 – 10x + 5
Expand
Remove brackets
and combine

76
If the binomials are in x and y, then the products consist of
the x2, xy and y2 terms. That is,
Example D. Expand.
(3x – 4y)(x + 5y)
= 3x2 + 15xy – 4yx – 20y2
= 3x2 + 11xy – 20y2
(#x + #y)(#x + #y) = #x2 + #xy + #y2
The FOIL method is still applicable in this case.
B. Expand and simplify.
Special Binomial Operations
1. (x + 5)(x + 7) 2. (x – 5)(x + 7)
3. (x + 5)(x – 7) 4. (x – 5)(x – 7)
5. (3x – 5)(2x + 4) 6. (–x + 5)(3x + 8)
7. (2x – 5)(2x + 5) 8. (3x + 7)(3x – 7)
Exercise. A. Expand by FOIL method first. Then do them by
inspection.
9. (–3x + 7)(4x + 3) 10. (–5x + 3)(3x – 4)
11. (2x – 5)(2x + 5) 12. (3x + 7)(3x – 7)
13. (9x + 4)(5x – 2) 14. (–5x + 3)(–3x + 1)
15. (5x – 1)(4x – 3) 16. (6x – 5)(–2x + 7)
17. (x + 5y)(x – 7y) 18. (x – 5y)(x – 7y)
19. (3x + 7y)(3x – 7y) 20. (–5x + 3y)(–3x + y)
21. –(2x – 5)(x + 3) 22. –(6x – 1)(3x – 4)
23. –(8x – 3)(2x + 1) 24. –(3x – 4)(4x – 3)
25. (3x – 4)(x + 5) + (2x – 5)(x + 3)
26. (4x – 1)(2x – 5) + (x + 5)(x + 3)
27. (5x – 3)(x + 3) + (x + 5)(2x – 5)
28. (3x – 4)(x + 5) – (2x – 5)(x + 3)
29. (4x – 4)(2x – 5) – (x + 5)(x + 3)
30. (5x – 3)(x + 3) – (x + 5)(2x – 5)
31. (2x – 7)(2x – 5) – (3x – 1)(2x + 3)
32. (3x – 1)(x – 7) – (x – 7)(3x + 1)
33. (2x – 3)(4x + 3) – (x + 2)(6x – 5)
34. (2x – 5)2 – (3x – 1)2
35. (x – 7)2 – (2x + 3)2
36. (4x + 3)2 – (6x – 5)2

77
The two binomials (A + B) and (A – B) are said to be the
conjugate of each other.
There are some important patterns in multiplying expressions
that it is worthwhile to memorize.
I. Difference of Squares Formula
(A + B)(A – B) = A2 – B2
To verify this :
(A + B)(A – B) = A2 – AB + AB – B2
= A2 – B2
Conjugate Product Difference of Squares
4– 6 Multiplication Formulas
For example, the conjugate of (3x + 2) is (3x – 2), and
the conjugate of (2ab – c) is (2ab + c).
Note: The conjugate is different from the opposite.
The opposite of (3x + 2) is (–3x – 2).
Example A. Expand.
a. (3x + 2)(3x – 2) = (3x)2 – (2)2 = 9x2 – 4
(A + B)(A – B) = A2 – B2
b. (2xy – 5z2)(2xy + 5z2)
= (2xy)2 – (5z2)2
= 4x2y2 – 25z4
Here are some examples of squaring: (3x)2 = 9x2,
(2xy)2 = 4x2y2, and (5z2)2 = 25z4.
II. Square Formulas
(A + B)2 = A2 + 2AB + B2
(A – B)2 = A2 – 2AB + B2
We may check this easily by multiplying,
(A + B)2 = (A + B)(A + B) = A2 + AB + BA + B2 = A2 + 2AB + B2
We say that “(A + B)2 is A2, B2, plus twice A*B”,
and “(A – B)2 is A2, B2, minus twice A*B”.
Example B.
a. (3x + 4)2 = (3x)2 + 2(3x)(4) + 42 = 9x2 + 24x + 16
(A + B)2 = A2 + 2AB + B2
b. (3a – 5b)2 = (3a)2 – 2(3a)(5b) + (5b)2
= 9a2 – 30ab + 25b2
III. Some Applications of the Formulas
We can use the above formulas to help us multiply.
The conjugate formula
(A + B)(A – B) = A2 – B2
may be used to multiply two numbers of the forms
(A + B) and (A – B) where A2 and B2 can be calculated easily.
Example C. Calculate. Use the conjugate formula.
a. 51*49 = (50 + 1)(50 – 1) = 502 – 12 = 2,500 – 1 = 2,499
b. 52*48 = (50 + 2)(50 – 2) = 502 – 22 = 2,500 – 4 = 2,496
c. 63*57 = (60 + 3)(60 – 3) = 602 – 32 = 3,600 – 9 = 3,591

78
Example D. Calculate. Use the squaring formulas.
a. 512 = (50 + 1)2 = 502 + 12 + 2(50)(1)
= 2,500 + 1 + 100
= 2,601
“(A + B)2 is A2, B2, plus twice A*B”,
“(A – B)2 is A2, B2, minus twice A*B”.
The Squaring Formulas.
b. 492 = (50 – 1)2 = 502 + 12 – 2(50)(1)
= 2,500 + 1 – 100
= 2,401
b. (50½) 2 = (50 + ½ )2 = 502 + ½ 2 + 2 (½) (50)
= 2,500 + 1/4 + 50
= 2,550¼
Exercise. A. Calculate. Use the conjugate formula.
Multiplication Formulas
1. 21*19 2. 31*29 3. 41*39 4. 71*69
5. 22*18 6. 32*28 7. 52*48 8. 73*67
B. Calculate. Use the squaring formula.
9. 212 10. 312 11. 392 12. 692
13. 982 14. 30½2 15. 100½2 16. 49½2
18. (x + 5)(x – 5) 19. (x – 7)(x + 7)
20. (2x + 3)(2x – 3) 21. (3x – 5)(3x + 5)
C. Expand.
22. (7x + 2)(7x – 2) 23. (–7 + 3x )(–7 – 3x)
24. (–4x + 3)(–4x – 3) 25. (2x – 3y)(2x + 3y)
26. (4x – 5y)(5x + 5y) 27. (1 – 7y)(1 + 7y)
28. (5 – 3x)(5 + 3x) 29. (10 + 9x)(10 – 9x)
30. (x + 5)2 31. (x – 7)2
32. (2x + 3)2 33. (3x + 5y)2
34. (7x – 2y)2 35. (2x – h)2
In this section, we carry out the operations in a vertical
format. To add and subtract in the vertical format, line up
like-terms vertically then add or subtract.
Example A.
a. Do vertically (–4x + 5) + (6x – 8)
–4x + 5
6x – 8+)
2x – 3
4–7 Polynomial Operations II
b. Do vertically (4x2 + 8x + 5) + (–6x2 – 8x – 3 )
4x2 + 8x + 5
–6x2 – 8x – 3
–2x2 + 0 + 2
Hence (4x2 + 8x + 5) + (–6x2 – 8x – 3 )
= –2x2 + 2
+)
Hence (–4x + 5) + (6x – 8)
= 2x – 3

79
Example B.
a. Do vertically (–4x + 5) – (6x – 8)
–4x + 5
6x – 8–)  change signs and add
– +
–10x +13
Hence (–4x + 5) – (6x – 8)
= –10x + 13
Vertical subtraction is used often. For examples,
4x
–6x
–4x
6x
–4x
–6x
4x
6x–) –) –) –)
–2x 10x –10x 2x
One way to help the subtraction is to change the signs of
terms subtracted then add.
Example C.
Do vertically (4x2 + 5) – (–6x2 + 8x – 3 )
4x2 + 5
–6x2 + 8x – 3–)
10x2 – 8x + 8
Hence (4x2 + 5) – (–6x2 + 8x – 3 )
= 10x2 – 8x + 8
+ – +
To multiply vertically, line up the like-terms in the product.
Example D. Do vertically (–4x + 5)(2x – 3)
–4x + 5
x) 2x – 3
–8x2 + 12x
+ 10x – 15+)
–8x2 + 22x – 15
Hence (–4x + 5)(2x – 3) = –8x2 +22x – 15
Example D. Do vertically (x2 – 4x – 3)(3x + 2)
x2 – 4x – 3
x) 3x + 2
3x3 –12x2– 9x
2x2 – 8x – 6+)
3x3 –10x2 –17x –6
Hence (x2 – 4x – 3)(3x + 2)
= 3x3 – 10x2 – 17x – 6
We may switch the order of multiplication.
Exercise. Do problem 18 – 29 from the last section using
the vertical format.

80
Chapter 5 – Factoring and Polynomial Equations
Example A. Factor 12 completely.
12 = 3 * 4 = 3 * 2 * 2
not prime factored completely
To factor means to rewrite a quantity as a product (in a
nontrivial way). A quantity x that can‟t be written as product
besides as 1*x is said to be prime. To factor completely
means each factor in the product is prime.
5–1 Factoring Out GCF
Example B.
a. Sincer 6 = 2*3, 15 = 3*5, 3 is a common factor.
b. The common factor of 4ab, 6a are 2, a, 2a.
c. The common factors of 6xy2, 15x2y2 are 3, x, y2, xy2, ..
d. The common factor of a(x+y), b(x+y) is (x+y).
A common factor of two or more quantities is a factor
belongs to all the quantities.
The greatest common factor (GCF) is the common factor
that has the largest coefficient and highest degree of each
factor among all common factors.
Example C. Find the GCF of the given quantities.
a. GCF{24, 36} = 12.
b. GCF{4ab, 6a} = 2a.
c. GCF {6xy2, 15 x2y2} = 3xy2.
d. GCF{x3y5, x4y6, x5y4} = x3y4.
The Extraction Law
Distributive law interpreted backward gives the Extraction Law,
that is, common factors may be extracted from sums or
differences.
AB AC  A(B C)
This procedure is also called “factoring out a common factor”.
To factor, the first step always is to factor out the GCF.
We may pull out common factors that are ( )'s.
Example E. Factor
a. a(x + y) – 4(x + y)
Pull out the common factor (x + y)
a(x + y) – 4(x + y) = (x + y)(a – 4)
b. Factor (2x – 3)3x – 2(2x – 3)
Pull out the common factor (2x – 3),
(2x – 3)3x – 2(2x – 3) = (2x – 3)(3x – 2)
(the GCF is y)
(the GCF is 2a)
(the GCF is 6x2y2)
Example D. Factor out the GCF.
a. xy – 4y = y(x – 4)
b. 4ab + 6a = 2a(2b) + 2a(3) = 2a(2b + 3)
c. 12x2y3 + 6x2y2 = 6x2y2(2y) + 6x2y2(1) = 6x2y2(2y + 1)

81
Example F. Pull out the negative sign from –2x + 5.
–2x + 5 = –(2x – 5)
Sometime we have to use factor twice.
Example G. Factor by pulling out twice.
a. y(2x – 5) – 2x + 5
= y(2x – 5) – (2x – 5)
= (2x – 5)(y – 1)
b. 3x – 3y + ax – ay Group them into two groups.
= (3x – 3y) + (ax – ay) Factor out the GCF of each group.
= 3(x – y) + a(x – y) Pull the factor (x – y) again.
= (3 + a)(x – y)
We sometimes pull out a negative sign from an expression.
Exercise. A. Find the GCF of the listed quantities.
Factoring Out GCF
1. {4, 6 } 2. {12, 18 } 3. {32, 20, 12 } 4. {25, 20, 30 }
5. {4x, 6x2 } 6. {12x2y, 18xy2 }
7. {32A2B3, 20A3B3, 12 A2B2}
8. {25x7y6z6, 20y7z5x6, 30z8x7y6 }
B. Factor out the GCF.
9. 4 – 6y 10. 12x + 18y 11. 32A + 20B – 12C
12. 25x + 20y – 30 13. –4x + 6x2
14. –12x2y – 18xy2 15. 32A2B3 – 20A3B3 – 12A2B2}
16. 25x7y6z6 – 20y7z5x6 + 30z8x7y6
17. 4x4 – 8x3 + 2x2 18. 20x4 – 5x2
19. x(x – 2) + 3(x – 2) 20. 4x(2x – 3) – 5(2x – 3)
C. Factor out the “–”.
21. –2y + 4 22. –3x + 18 23. –5x + 15 24. –8x + 16
D. Factor, use grouping if it‟s necessary.
25. y2 – 2y + 3y – 6 26. x2 + 3x + 6x + 18
27. y2 – 2y – 3y + 6 28. x2 + 3x – 6x – 18
29. y2 – y + 4y – 4 30. x2 – 5x – 2x + 10
31. 2y2 – y – 6y + 3 32. 3x2 + 2x – 6x – 4
33. 4x2 + 6x – 6x – 9 34. –3x2 + 4x – 6x + 8
35. –5y2 + 10y – 3y + 6 36. –x2 + 3x – 7x + 21
37. 2y2 – xy – 6xy + 3x2 38. 3x2 + 2xy – 6xy – 4y2
39. –5x2 + 2xy – 20xy + 8y2 40. –14x2 + 21xy – 8xy + 12y2

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5–2 Factoring Trinomials I
To factor x2 + bx + c, we note hat
if (x + u)(x + v)
= x2 + ux + vx + uv
= x2 + (u + v)x + uv
= x2 + bx + c,
we need to have u and v where uv = c, and u + v = b. If this
can‟t be done, then the trinomial is prime (not factorable).
Trinomials (three-term) are polynomials of the form
ax2 + bx + c where a, b, and c are numbers. The product of
two binomials is a trinomials:
(#x + #)(#x + #)  ax2 + bx + c
Hence, to factor a trinomial, we write the trinomial as a
product of two binomials, if possible, that is:
ax2 + bx + c  (#x + #)(#x + #)
We start with the case where a = 1, or trinomials
of the form x2 + bx + c.
Example A.
a. Factor x2 + 5x + 6
We want (x + u)(x + v) = x2 + 5x + 6, so we need u and v
where uv = 6 and u + v = 5.
Since 6 = (1)(6) = (2)(3) = (-1)(-6) = (-2)(-3) and 2 + 3 = 5,
so x2 + 5x + 6 = (x + 2)(x + 3).
c. Factor x2 + 5x – 6
We want (x + u)(x + v) = x2 + 5x – 6, so we need uv = –6 and
u + v = 5.
Since -6 = (–1)(6) = (1)(–6) = (–2)(3) =(2)(–3) and –1 + 6 = 5,
so x2 + 5x – 6 = (x – 1)(x + 6).
b. Factor x2 – 5x + 6
We want (x + u)(x + v) = x2 – 5x + 6, so we need u and v
where uv = 6 and u + v = –5.
Since 6 = (1)(6) = (2)(3) = (-1)(-6) = (-2)(-3) and –2 – 3 = –5,
so x2 – 5x + 6 = (x – 2)(x – 3).
{
Observations About Signs
Given that x2 + bx + c = (x + u)(x + v) so that uv = c,
we observe the following.
1. If c is positive, then u and v have same sign.
In particular,
if b is also positive, then both are positive.
if b is negative, then both are negative.
From the examples above
x2 + 5x + 6 = (x + 2)(x + 3)
x2 – 5x + 6 = (x – 2)(x – 3)
2. If c is negative, then u and v have opposite signs. The
one with larger absolute value has the same sign as b.
From the example above
x2 – 5x – 6 = (x – 6)(x + 1)

83
Example B.
a. Factor x2 + 4x – 12
We need u and v having opposite signs such that uv = –12,
u + v = +4. Since -12 = (-1)(12) = (-2)(6) = (-3)(4)…
They must be –2 and 6 hence x2 + 4x – 12 = (x – 2)(x + 6).
b. Factor x2 – 8x – 12
We need u and v such that uv = –12, u + v = –8 with
u and v having opposite signs. This is impossible.
Hence x2 – 8x – 12 is prime.
Factoring Trinomials I
Exercise. A. Factor. If it‟s prime, state so.
1. x2 – x – 2 2. x2 + x – 2 3. x2 – x – 6 4. x2 + x – 6
5. x2 – x + 2 6. x2 + 2x – 3 7. x2 + 2x – 8 8. x2 – 3x – 4
9. x2 + 5x + 6 10. x2 + 5x – 6
13. x2 – x – 20
11. x2 – 5x – 6
12. x2 – 5x + 6
17. x2 – 10x – 24
14. x2 – 8x – 20
15. x2 – 9x – 20 16. x2 – 9x + 20
18. x2 – 10x + 24 19. x2 – 11x + 24 20. x2 – 11x – 24
21. x2 – 12x – 36 22. x2 – 12x + 36 23. x2 – 13x – 36
24. x2 – 13x + 36
B. Factor. Factor out the GCF, the “–”, and arrange the
terms in order first if necessary.
29. 3x2 – 30x – 7227. –x2 – 5x + 14 28. 2x3 – 18x2 + 40x
30. –2x3 + 20x2 – 24x
25. x2 – 36 26. x2 + 36
31. –2x4 + 18x2
32. –3x – 24x3 + 22x2 33. 5x4 + 10x5 – 5x3
Factoring Trinomials I
35. –3x3 – 30x2 – 48x34. –yx2 + 4yx + 5y
36. –2x3 + 20x2 – 24x
40. 4x2 – 44xy + 96y2
37. –x2 + 11xy + 24y2
38. x4 – 6x3 + 36x2 39. –x2 + 9xy + 36y2
C. Factor. Factor out the GCF, the “–”, and arrange the
terms in order first.
D. Factor. If not possible, state so.
41. x2 + 1 42. x2 + 4 43. x2 + 9 43. 4x2 + 25
44. What can you conclude from 41–43?

84
5–3 Factoring Trinomials II
Now let‟s try to factor trinomials of the form ax2 + bx + c.
We‟ll give two methods. One is short but not reliable.
The second one takes more steps but gives definite answers.
For this method, we need to find four numbers that fit certain
descriptions. The following are examples of the task to be
accomplished.
Example A. Let {1, 3} and {1, 2} be two pairs of numbers.
Is it possible to split the {1, 2 }, put them in the boxes that
makes the equality true?
a. 1* (± ) + 3*(± ) = 5.
Yes, 1* (2) + 3 * (1) = 5
b. 1* (± ) + 3* (± ) = –5.
Yes, 1* (1) + 3* (–2) = –5 or 1* (–2) + 3* (–1) = –5
c. 1* (± ) + 3* (± ) = 8.
No, since the most we can obtain is 1* (1) + 3* (2) = 7.
Reversed FOIL Method
3(±# ) +1(±#) = 5 where the #‟s are 1 and 2.
Since 3(1) +1(2) = 5, we see that
3x2 + 5x + 2 = (3x + 2)(1x + 1).
5x
(Reversed FOIL Method)
Let‟s see how the above examples are related to factoring.
Example B. Factor 3x2 + 5x + 2.
The only way to get 3x2 is (3x ±#)(1x ±#).
The #‟s must be 1 and 2 to get the constant term +2.
We need to place 1 and 2 as the #'s so the product will
yield the correct middle term +5x.
That is, (3x ±#)(1x ±#) must yields +5x, or that
3(±# ) + 1(±# ) = –7.
It's 3(–2) + 1(–1) = –7.
So 3x2 – 7x + 2 = (3x –1)(1x – 2)
Example D. Factor 3x2 + 5x – 2.
We start with (3x ±#)(1x ±#).
We need to fill in 1 and 2 so that
3(±# ) + 1(±# ) = +5.
It is 3(+2) + 1(–1) = +5.
So 3x2 + 5x + 2 = (3x –1)(1x + 2)
Since c is negative, they must have opposite signs .
Example C. Factor 3x2 – 7x + 2.
We need to fill in 1 and 2 as #'s so that

85
Example E. Factor 3x2 + 8x + 2.
We need to fill in 1&2 so that
3(±# ) + 1(±# ) = +8.
This is impossible. Hence the expression is prime.
3(±# ) + 1(±# ) = +11. It can't be 2&2.
If both the numbers a and c in ax2 + bx + c have many factors
then there are many possibilities to check.
Example F. Factor 3x2 + 11x – 4.
We start with (3x ±#)(1x ±#). Since 4 = 2(2) = 1(4),
we need to fill in 2&2 or 1&4 so that
Try 1&4, it is
3(+4) + 1(–1) = +11.
So 3x2 + 11x – 4 = (3x – 1)(1x + 4).
Example G. Factor 12x2 – 5x – 3.
Since 3 must be 3(1), we need to find two numbers such
that (#)(#) = 12 and that
So 12x2 – 5x – 3 = (3x + 1)(4x – 3).
It's not necessary to always start with ax2. If c is a prime
number, we start with c.
(±#)(±3) + (±#)(±1) = – 5.
12 = 1(12) = 2(6) = 3(4)
1&12 and 2&6 can be quickly eliminated.
We get (3)(–3) + (4)(+1) = – 5.
Remark:
In the above method, finding
(#)(±#) + (#)( ±#) = b
does not guarantee that the trinomial will factor. We have to
match the sign of c also.
Example H. Factor 3x2 – 7x – 2 .
We start with (3x ±#)(1x ±#). We find that:
3(–2) + 1(–1) = –7.
But this won't work since (–2)(–1) = 2 = c.
In fact this trinomial is prime.
Example I: Factor 1x2 + 5x – 6 .
We have:
1(+3) + 1(+2) = +5
The one that works is x2 + 5x – 6 = (x + 6)(x – 1).
1(+6) + 1(–1) = +5
There might be multiple matchings for
(#)(±#) + (#)( ±#) = b
make sure you chose the correct one, if any.

86
Finally, before starting the reverse-FOIL procedure
1. make sure the terms are arranged in order.
2. if there is any common factor, pull out the GCF first.
3. make sure that x2 is positive, if not, factor out the negative
sign first.
Example J. Factor –x3 + 3x + 2x2
–x3 + 3x + 2x2 Arrange the terms in order
= –x3 + 2x2 + 3x Factor out the GCF
= – x(x2 – 2x – 3)
= – x(x – 3)(x + 1)
Exercise. A. Factor completely. If it‟s prime, state so.
1. 3x2 – x – 2 2. 3x2 + x – 2 3. 3x2 – 2x – 1
4. 3x2 + 2x – 1 5. 2x2 – 3x + 1 6. 2x2 + 3x – 1
8. 2x2 – 3x – 27. 2x2 + 3x – 2
15. 6x2 + 5x – 6
10. 5x2 + 9x – 2
9. 5x2 – 3x – 2
12. 3x2 – 5x + 211. 3x2 + 5x + 2
14. 6x2 – 5x – 613. 3x2 – 5x + 2
16. 6x2 – x – 2 17. 6x2 – 13x + 2 18. 6x2 – 13x + 2
19. 6x2 + 7x + 2 20. 6x2 – 7x + 2 21. 6x2 – 13x + 6
22. 6x2 + 13x + 6 23. 6x2 – 5x – 4 24. 6x2 – 13x + 8
25. 6x2 – 13x – 8
Factoring Trinomials II
25. 4x2 – 9 26. 4x2 – 49
27. 25x2 – 4 28. 4x2 + 9 29. 25x2 + 9
30. – 6x2 – 5xy + 6y2 31. – 3x2 + 2x3– 2x 32. –6x3 – x2 + 2x
33. –15x2 – 25x2 – 10x 34. 12x2y2 –14x2y2 + 4xy2

87
5–4 Factoring Trinomials III
Example I. Use the b2 – 4ac to see if the trinomial is
factorable. If it is, factor it.
a. 3x2 – 7x – 2
b2 – 4ac = (–7)2 – 4(3)(–2) = 49 + 24 = 73 is not a square,
hence it is prime.
Theorem: If b2 – 4ac = 0, 1, 4, 9, 16, 25, 36, .. i.e. is a
squared number, then the trinomial is factorable.
Otherwise, it is not factorable.
In this section we give a formula that enables us to tell if a
trinomial is factorable or not.
Also we give the ac-method for factoring.
b. 3x2 – 7x + 2
b2 – 4ac = (–7)2 – 4(3)(2) = 49 – 24 = 25 which is a squared
number, hence it is factorable.
In fact 3x2 – 7x + 2 = (3x – 1)(x – 2)
The ac-Method
The product of two binomials has four terms. We may use
grouping-method to work backward to obtain the two
binomials.
But how do we know to write x2 – x – 6 as x2 – 3x + 2x – 6 in
the first place? This leads to the ac-method. The ac-method
tells us how to rewrite the trinomial so we may use the
grouping method. If it isn't possible to rewrite the trinomial
according to the ac-method, then the trinomial is prime.
Example A. Factor x2 – x – 6 by grouping.
x2 – x – 6
= x2 – 3x + 2x – 6 Put them into two groups
= (x2 – 3x) + (2x – 6) Take out the common factors
= x(x – 3) + 2(x – 3) Take out the common (x – 3)
= (x – 3)(x + 2)

88
Example B. Factor 3x2 – 4x – 20 using the ac-method.
a = 3, c = –20. Hence ac = 3(–20) = –60. We need two
numbers u and v such that uv = –60 and u + v = –4.
By searching, we get that 6(–10) = –60 and 6 + (–10) = –4.
Write 3x2 – 4x – 20
= 3x2 + 6x –10x – 20 Put in two groups,
= (3x2 + 6x ) + (–10x – 20) Pull out common factor
= 3x(x + 2) – 10 (x + 2) Pull out common factor
= (3x – 10)(x + 2)
ac-Method: We assume that there is no common factor for
the trinomial ax2 + bx + c.
1. Calculate ac, and find two numbers u and v such that
uv is ac, and u + v = b.
2. Write ax2 + bx + c as ax2 + ux + vx +c
3. Use the grouping method to factor (ax2 + ux) + (vx + c)
If step 1 can‟t be done, then the expression is prime.
Example C. Factor 3x2 – 6x – 20 using the ac-method.
a = 3, c = –20, hence ac = 3(-20) = –60.
Need two numbers u and v such that
uv = –60 and u + v = –6.
After searching all possibilities we found that it's impossible.
Hence 3x2 – 6x – 20 is prime.
1. 3x2 – x – 2 2. 3x2 + x – 2 3. 3x2 – 2x – 1
4. 3x2 + 2x – 1 5. 2x2 – 3x + 1 6. 2x2 + 3x – 1
8. 2x2 – 3x – 27. 2x2 + 3x – 2
15. 6x2 + 5x – 6
10. 5x2 + 9x – 2
9. 5x2 – 3x – 2
12. 3x2 – 5x + 211. 3x2 + 5x + 2
14. 6x2 – 5x – 613. 3x2 – 5x + 2
16. 6x2 – x – 2 17. 6x2 – 13x + 2 18. 6x2 – 13x + 2
19. 6x2 + 7x + 2 20. 6x2 – 7x + 2 21. 6x2 – 13x + 6
22. 6x2 + 13x + 6 23. 6x2 – 5x – 4 24. 6x2 – 13x + 8
25. 6x2 – 13x – 8
Factoring Trinomials III
25. 4x2 – 9 26. 4x2 – 49
27. 25x2 – 4 28. 4x2 + 9 29. 25x2 + 9
30. – 6x2 – 5xy + 6y2 31. – 3x2 + 2x3– 2x 32. –6x3 – x2 + 2x
33. –15x2 – 25x2 – 10x 34. 12x2y2 –14x2y2 + 4xy2
Exercise A. Use the ac–method, factor the trinomial or
demonstrate that it‟s not factorable.

89
We may evaluate expressions with numbers as inputs.
5– 5 Substitution and Factoring Formulas
For example, evaluate A2 – B2 with A = 2 and B = 1 as inputs,
we have 3. We extend this to evaluating expressions with
symbols as inputs. This is called "substitution". In such
problems always simplify the answers as much as possible.
Example A. Evaluate A2 – B2 with A = (x + 2) and B = (2x).
Replace A by (x + 2), and B by (2x) we have
(x + 2)2 – (2x)2
= x2 + 4x + 4 – 4x2
= –3x2 + 4x + 4
Vice versa, for some outputs we can easily identify the inputs
especially when the formula is given.
Example B. With inputs for A and B, evaluate A2 – B2
a. we have 4x2 – 9y2, what are A and B?
Match the A2 = 4x2, we‟re asking ( ? )2 = 4x2, so A = 2x.
Similarly we conclude that B = 3y.
c. we have x2 + 2xy + y2 – 9, what are A and B?
b. we have 25x2y4 – 1, what are A and B?
Match the A2 = 25x2y4 and B2 = 1, we have that
A = 5xy2 and B = 1.
so x2 + 2xy + y2 – 9 = (x + y)2 – 32.
Hence A = (x + y) and B = 3.
a. we have x3 – 125, what are A and B?
Match the A3 = x3, we have A = x.
Similarly we‟ve B3 = 125 so B = 5.
b. we have 8y3 – (2 – y)3 , what are A and B?
Match the A3 =8y3, we have A = 2y.
By inspection, we have B = (2 – y).
Recall the conjugate product formula
(A – B)(A + B) = A2 – B2.
Factor x2 + 2xy + y2 we have (x + y)(x + y) = (x + y)2,
Example C. Evaluate A3 – B3 with inputs A and B
Example D.
a. x2 – 25
= (x)2 – (5)2
= (x + 5)(x – 5)
b. 4x2 – 9y2
= (2x)2 – (3y)2
= (2x – 3y)(2x + 3y)
c. 25A2B4 – 1
= (5AB2)2 – (1)2
= (5AB2 – 1)(5AB2 + 1)
The reverse of the conjugate product formula is the
factoring formula for difference of squares :
A2 – B2 = (A – B) (A + B)
As suggested by the name, we use this formula when
it is the subtraction of two perfect square terms.

90
d. (A + 2B)2 – 16
= (A + 2B)2 – (4)2
= ((A + 2B) – 4)((A + 2B) + 4)
= (A + 2B – 4)(A + 2B + 4)
Sum and Difference of Cubes
Both the difference and the sum of cubes may be factored.
If we try to factor A3 – B3, we might guess that
A3 – B3 = (A – B)(A2 + B2)
But this does not work because we have the inner
and outer products –BA2 and AB2 remaining in the expansion
as shown above.
Caution: The expressions
A2 + B2 or – A2 – B2
are not factorable!
But if we put in the adjustment +AB, then all the cross-
products with A and B are eliminated. That is
(A – B)(A2 + AB + B2)
= A3 + A2B + AB2 – BA2 – B2A – B3
= A3 – B3
Hence A3 – B3 = (A – B)(A2 + AB + B2)
Through a similar argument we have that
A3 + B3 = (A + B)(A2 – AB + B2)
We write these two formulas together as
A3 B3 = (A B)(A2 AB + B2)+– +–+–
Note that the signs are reversed for the adjustments in
these formulas.
Example E.
a. 8x3 – 27 (Note that both 8 and 27 are cubes)
= (2x)3 – (3)3
= (2x – 3)((2x)2 + (2x)(3) +(3)2)
= (2x – 3)(4x2 + 6x + 9)
b. 64A3 + 125 (both 64 and 125 are cubes)
= (4A)3 + (5)3
= (4A + 5)((4A)2 – (4A)(5) +(5)2)
= (4A + 5)(16A2 – 20A + 25)
Summary on Factoring
I. Always factor out the GCF first.
II. Make sure the leading term is positive and all terms are
arranged in order.
III. If left with a trinomial, use reverse-FOIL or ac-method.
IV. If it is a two-term expression, try factoring by formula.
V. If it is a four term expression, try the grouping-method.

91
Example F. Factor completely.
A. 20A3 – 45A Take out the common factor
= 5A(4A2 – 9) Difference of square
= 5A(2A – 3)(2A + 3)
B. –20A3 + 45A2 – 10A Take out the common factor and -1
= –5A(4A2 – 9A + 2)
Exercise A. Simplify with the given substitution.
Simplify your answers when possible.
For 1- 8, Simplify A2 – B2 with the given substitution.
1. A = x, B = 1 2. A = 3, B = 2y
3. A = 2xy, B = 1 4. A = x + y, B = 2
5. A = (x – 3), B = x 6. A = 2x, B = (x – 1)
7. A = (x – 3), B = (x – 1) 8. A = (2x + 1), B = (2x – 1)
For 9-14, simplify A3 – B3 with the given substitution.
9. A = x, B = 2 10. A = 3x, B = 2
11. A = 2x, B = 3y 12. A = 1, B = 2xy
13. A = 4x, B = 5 14. A = 6x, B = 5y
15. Simplify x2 – 2x + 10 with the substitution x = (2 + h)
16. Simplify x2 + 3x – 4 with the substitution x = (–3 + h)
Substitution and Factoring Formulas
B. Identify the substitution for A and B then factor the
expression using the pattern A2 – B2 = (A – B)(A + B).
17. x2 – 1 18. 9 – 4y2
23. 4x2y2 – 1
26. x2 + 2xy + y2 – 4
21. 25x2 – 64 22. 9 – 49y2
25. 4 – 81x2y2
24. 98x2 – 8
19. x2 – 9 20. 25 – y2
C. Identify the substitution for A and B then factor the
expression using the pattern A3 ± B3 = (A ± B)(A2 AB + B2).+
29. 8x2z – 92z
27. 50x2 – 18y2 28. 36x3 – 25x
30. x2 – y2 – 2y – 1
31. x3 – 1 32. x3 – 8
37. x3 – 125
36. 27 – 8y334. x3 + 8 35. 64 + y3
33. 27 – y3
38. 27x3 – 64 39. x3 – 1000

92
5–6 Applications of Factoring
There are many applications of the factored forms of
polynomials. Following are some of them:
1. to evaluate polynomials,
2. to determine the signs of the outputs,
3. most importantly, to solve polynomial-equations.
Evaluating Polynomials
Example A. Evaluate x2 – 2x – 3 if x = 7
a. without factoring. b. by factoring it first.
We get
72 – 2(7) – 3
= 49 – 14 – 3
= 32
x2 – 2x – 3 = (x – 3)(x+1)
We get
(7 – 3)(7 + 1)
= 4(8)
= 32
Often it is easier to evaluate polynomials in the factored form.
Example B. Evaluate 2x3 – 5x2 + 2x
for x = -2, -1, 3 by factoring it first.
2x3 – 5x2 + 2x = x(2x2 – 5x + 2)
= x(2x – 1)(x – 2)
For x = -2:
(-2)[2(-2) – 1] [(-2) – 2] = -2 [-5] [-4] = -40
For x = -1:
(-1)[2(-1) – 1] [(-1) – 2] = -1 [-3] [-3] = -9
For x = 3:
3 [2(3) – 1] [(3) – 2] = 3 [5] [1] = 15
Determine the signs of the outputs
Often we only want to know the sign of the output, i.e.
whether the output is positive or negative. It is easy to do this
using the factored form.
Example C. Determine the outcome is + or – for x2 – 2x – 3
if x = -3/2, -1/2.
Factor x2 – 2x – 3 = (x – 3)(x + 1)
Hence for x = -3/2:
(-3/2 – 3)(-3/2 + 1) is (–)(–) = + . So the outcome is positive.
And for x = -1/2:
(-1/2 – 3)(-1/2 + 1) is (–)(+) = – . So the outcome is negative.
Solving Equations
The most important application for factoring is to solve
polynomial equations. These are equations of the form
polynomial = polynomial
To solve these equations, we use the following obvious fact.
Fact: If A*B = 0,
then either A = 0 or B = 0
For example, if 3x = 0, then x must be equal to 0.

93
Example D.
a. If 3(x – 2) = 0, then (x – 2) = 0, so x must be 2.
To solve polynomial equation,
1. set one side of the equation to be 0, move all the terms to
the other side.
2. factor the polynomial,
3. get the answers.
b. If (x + 1)(x – 2) = 0,
then either (x + 1) = 0 or (x – 2) = 0,
x = –1 or x = 2
Example E. Solve for x
a. x2 – 2x – 3 = 0 Factor
(x – 3)(x + 1) = 0
Hence x – 3 = 0 or x + 1 = 0
x = 3 or x = -1
b. 2x(x + 1) = 4x + 3(1 – x) Expand
2x2 + 2x = 4x + 3 – 3x
2x2 + 2x = x + 3 Set one side 0
2x2 + 2x – x – 3 = 0
2x2 + x – 3 = 0 Factor
(2x + 3)(x – 1) = 0 Get the answers
2x + 3 = 0
2x = -3
x = -3/2
or x – 1 = 0
x = 1
c. 8x(x2 – 1) = 10x Expand
8x3 – 8x = 10x Set one side 0
8x3 – 8x – 10x = 0
8x3 – 18x = 0 Factor
2x(2x + 3)(2x – 3) = 0 Get the answers
x = 0 or 2x + 3 = 0 or 2x – 3 = 0
2x = -3 2x = 3
x = -3/2 x = 3/2
Exercise A. Use the factored form to evaluate the following
expressions with the given input values.
Applications of Factoring
1. x2 – 3x – 4, x = –2, 3, 5 2. x2 – 2x – 15, x = –1, 4, 7
3. x2 – 2x – 1, x = ½ ,–2, –½ 4. x3 – 2x2, x = –2, 2, 4
5. x3 – 4x2 – 5x, x = –4, 2, 6 6. 2x3 – 3x2 + x, x = –3, 3, 5
B. Determine if the output is positive or negative using the
factored form.
7. x2 – 3x – 4, x = –2½, –2/3, 2½, 5¼
8. –x2 + 2x + 8, x = –2½, –2/3, 2½, 5¼
9. x3 – 2x2 – 8x, x = –4½, –3/4, ¼, 6¼,
11. 4x2 – x3, x = –1.22, 0.87, 3.22, 4.01
12. 18x – 2x3, x = –4.90, –2.19, 1.53, 3.01
10. 2x3 – 3x2 – 2x, x = –2½, –3/4, ¼, 3¼,

94
C. Solve the following equations.
18. x2 – 3x = 10 20. x(x – 2) = 24 21. 2x2 = 3(x + 1) – 1
28. x3 – 2x2, x = –2, 2, 4
22. x2 = 4
25. 2x(x – 3) + 4 = 2x – 4
29. x3 – 2x2 – 8x = 0
31. 4x2 = x3
30. 2x2(x – 3) = 4x
26. x(x – 3) + x + 6 = 2x2 + 3x
13. x2 – 3x – 4 = 0 14. x – 2x – 15 = 0 15. x2 + 7x + 12 = 0
16. –x2 – 2x + 8 = 0 17. 9 – x2 = 0 18. 2x2 – x – 1 = 0
27. x(x + 4) + 9 = 2(2 – x)
23. 8x2 = 2 24. 27x2 – 12 = 0
32. 4x = x3 33. 4x2 = x4
34. 7x2 = –4x3 – 3x 35. 5 = (x + 2)(2x + 1)
36. (x – 1)2 = (x + 1)2 – 4 37. (x + 1)2 = x2 + (x – 1)2
38. (x + 2)2 – (x + 1)2= x2 39. (x + 3)2 – (x + 2)2 = (x + 1)2
Many physics formulas are 2nd degree.
If a stone is thrown straight up on Earth then
h = -16t2 + vt
5–7 2nd-Degree-Equation Word Problems
height = -16t2 + vt
after t secondswhere
h = height in feet
t = time in second
v = upward speed in feet per second
Example A. If a stone is thrown straight
up at a speed of 64 ft per second,
a. how high is it after 1 second?
t = 1, v = 64, so
h = -16(1)2 + 64(1)
h = -16 + 64 = 48 ft
c. What is the maximum height obtained?
Since it takes 4 seconds for the stone to fall back to the
ground, at 2 seconds it must reach the highest point.
Hence t = 2, v = 64, need to find h.
h = -16(2)2 + 64(2)
= - 64 + 128
= 64 (ft)
Therefore, the maximum height is 64 feet.
b. How long will it take for it to fall back to the ground?
To fall back to the ground means the height h is 0.
Hence
h = 0, v = 64, need to find t.
0 = -16 t2 + 64t
0 = -16t(t – 4)
t = 0 or t – 4 = 0 or t = 4
Therefore it takes 4 seconds.

95
If L and W are in a given unit, then A is in unit2.
For example: If L = 2 in, W = 3 in. then A = 2*3 = 6 in2
1 in
1 in
1 in2
2 in
3 in
6 in2
Area of a Rectangle
Given a rectangle, let
L = length of a rectangle
W = width of the rectangle,
the area A of the rectangle is
A = LW.
L
w A = LW
Formulas of area in mathematics also lead to 2nd degree
equations.
Example B. The length of a rectangle is 4 inches more than
the width. The area is 21 in2. Find the length and width.
Let x = width, then the length = (x + 4)
LW = A, so (x + 4)x = 21
x2 + 4x = 21
x2 + 4x – 21 = 0
(x + 7)(x – 3) = 0
x = - 7 or x = 3
Therefore, the width is 3 and the length is 7.
Area of a Parallelogram
A parallelogram is the area enclosed
by two sets of parallel lines. If we
move the shaded part as shown, we
get a rectangle. Hence the area A
of the parallelogram is A = BH.
H=height
B=base
Example C. The area of the parallelogram shown is 27 ft2.
Find x.
The area is (base)(height) or that
x(2x + 3) = 27
Area of a Triangle
Given the base (B) and the height (H) of a triangle as shown.
B
H
2x + 3
x
2x2 + 3x = 27
2x2 + 3x – 27 = 0 Factor
(2x + 9)(x – 3) = 0 x = –9/2, x = 3 ft
Take another copy and place it
above the original one as shown.
We obtain a parallelogram.
If A is the area of the triangle,
then 2A = HB or .A = BH
2

96
Example D. The base of a triangle is 3 inches shorter than
the twice of the height. The area is 10 in2. Find the base
and height.
Let x = height, then the base = (2x – 3)
Hence, use the 2nd version of the formula we have
20 = (2x – 3) x
20 = 2x2 – 3x
0 = 2x2 – 3x – 20
0 = (x – 4)(2x + 5)
x = 4 or x = -5/2
Example D. The base of a triangle is 3 inches shorter than
the twice of the height. The area is 10 in2. Find the base
and height.
Let x = height, then the base = (2x – 3)
Hence, use the formula 2A = BH
2*10 = (2x – 3) x
20 = 2x2 – 3x
0 = 2x2 – 3x – 20
0 = (x – 4)(2x + 5)
x = 4 or x = -5/2
Therefore the height is 4 in. and the base is 5 in.
2x– 3
x

97
Exercise A. Use the formula h = –16t2 + vt for the following
problems.
1. A stone is thrown upward at a speed of v = 64 ft/sec,
how long does it take for it‟s height to reach 48 ft?
how long does it take for it‟s height to reach 48 ft?
a. how long does it take for its height to reach 80 ft? Draw a
picture.
b. how long does it take for its height to reach the highest
point?
c. What is the maximum height it reached?
a. how long does it take for its height to reach 160 ft?
Draw a picture. How long does it take for its height to reach
the highest point and what is the maximum height it reached?
2nd-Degree-Equation Word Problems
B. Given the following area measurements, find x.
5.
8 ft2
x + 2
x
6.
12 ft2
x
(x – 1)
7.
x + 2
8.
12 ft2 x
(x + 4)
9.
24 ft2
(3x – 1)
x
10.
15 ft2x
18 ft2 x
(4x + 1)
2x + 1
11.
x5cm2
12.
x
2x – 3
9cm2
2x + 1
13.
x
18km2
14.
(x + 3)
(5x + 3)
24km2
15. 16.
16km2
2
x
x + 1
35km2
2
x
2x – 1

98
Solutions to the Odd Problems
The Basic Pg. 3 A. 1. 15 3. 18 5. 29 7. 28
9. 30 11. 48 13. 77 15. 107 17. 540
B. 1. 27 3. 25 5. 36 7. 49 9. 81 11. 1000
13. 100000 15. 1000000 17. 121
19.
2 3 2 4 2 2 3 5 2 2 2
4 3 2 2 3 2 2 4 3
2 , 2 , 3 2 , 2 , 2 3 , 6 2 , 3 , 2 , 2 3 , 5 3 ,
3 2 , 7 2 , 3 5 2 , 7 3 , 2 3 , 3 5 , 3 , 3 5 2
    
        
C. 21. 120 23. 540 25. 504 27. 2400 29. 1200 31. 16 33. 64 35.256
37. 1024 39. 243
Fractions and Mixed Numbers Pg. 5 A. 2/3, 2/3,5/3,4/3,5/7,3/2,5/4,2/3
B. 1.
1
4
2
3.
1
2
4
5.
1
3
12
7.
2
7
17
Multiplication and Division of Fractions Pg. 7
A. 1. .6 .10 .14 .18a b c d 3. .10 .15 .5 .27a b c d
5. .3 .15 .15 .54a b c d 7. .27 .24 .21 .14a b c d
9. .14 .44 .63 .30a b c d
B. 11. 30 13. 15 15. 24 17. 36 19. 24 21. 69
23. Adults Teens Children
Males 16 27 12
Female 32 9 12
Total 48 36 24
25. 76 ; 19/27
LCM and LCD Pg. 10 A. 1. a. 24 b. 18 c.12 d. 20 3. a.72 b. 90 c.180 5.
a.20 b.360 c. 144 7. a.72 b.120 c.240 9. a.360 b.720
B. 11.
4 18 20 9
24 24 24 24
13.
54 35 156 44
60 60 60 60
Addition and Subtraction of Fractions Pg. 13 A. 1. 2 3. 2 5.
2
5
7.
5
9

99
9.
7
9
11.
13
4
13.
52
5
15.
47
6
B. 17.
5
6
19.
13
6
21.
11
42
23.
53
44
C. 25.
17
12
27.
41
36
29.
29
48
31.
37
120
33.
17
36
35.
1
8
37.
7
144
39.
115
144
41.
5
6
43.
13
180
45.
121
24
Some Facts About Divisibility Pg. 16 A. 1. 106 can‟t be divided by 3 and 6 3. 402
can be divided by 3 and 6 5. 183420 can be divided by 3 and 6 7. 2040 can be divided
by 4 and 8 9. 1096 can be divided by 4 and 8 11. 1134, 183420,801171 can be
divided by 9 13. 1134,183420 can be divided by 18
Cross Multiplication Pg. 18 A. 1.
3
2
3.
9
4
5.
6
5
7.
21
20
9.
3
2
B. 11.
4
5
is more 13.
5
6
is more 15.
7
10
is more 17.
13
8
is more
C. 19.
1
6
21.
23
20
23.
3
10
25.
4
45
Signed Numbers Pg. 20 A. 1. 5 3. 76 5. -16 7. -60 9. -6 11. 11 13. -5 B.
15. 8 17. -21 19. -6 21.-12
Addition and Subtraction of Signed Numbers Pg. 22 A. 1. 10 3. -4 5. 0
7. -16 9. -32
B. 11. 4 13. -10 15. -6 17. 30 19. -2 C.
21. 1 23. -9 25. 28 27. 32 D.
29. 7 31. 13 33. 1 35. 150
Multiplication and Division of Signed Numbers Pg. 24 A. 1. 0 3. 0 5. 9 7.
0 B.
9. -6 11. -6 13. -8 15. -16 C.
17. -4 19. 3 21.
3
2
23.
3
2
25.
4
15
 27. -15
Order of Operations Pg. 26 A. 1. -9 3. -6 5. -27 B.
7. 7 9. 79 11. 11 13. 1 15. -25 17. -5 19. -39 21. -23 C.
23. 4 and -4 25. 16 and -16 27. 18 D.
29. -1 31. -14 33. -41 35. -10 E.
37. 81 39. -8 41. 33 43. 16 45. 98 47. 72

100
49. -15 51. 33 53. 6 55. -1/2
Variables and Evaluation Pg. 29 1. -6 3. 10 5. -8 7. 6 9. 9 11. 25 13.
-4 15. -81 17. -54 19. 2 21. 55 23. 4 25. 19 27. -29 29. 0 31. 3
33. -1 35. -2 37. -31 39. -8 41. 1/2 43. 1 45. 1 47. 4 49. -20
Expressions Pg. 33 A. 1. 8x 3. -8x 5. 8x + 4 7. -3x + 3 9. 2x-5 11.
6B-2A B.
13. 3x+15 15. 12x+20 17. 6x 19. 11x+42 21. -54x+76 23. 62x+68
25. 2B-2A 27. 9A-20B C.
29. 41x - 12 31. 13x – 32 33. – 40x + 69 35. 5x – 12 37. 7x/12 39. –29x/24 41.
240 bananas and 264 apples
Linear Equations I Pg. 36 A. 1. x=1 3. x=2 5. x=3/2 B.
7. x=3/2 9. x=3/5 11. x=-8 13. x=12 15. x=7 C.
17. x=-2 19.x=3 21. x=-1/3 23.. x=-2 D.
25. x=2/3 27. x=8/3 29. x=2 31. x=0 33. x=5 35. x=1
Linear Equations II Pg. 39 A. 1. x=3 3.x=4 5.x=-3 7.x=4/3 9.x=-7/3
B. 11.x=15/4 13.x=1/2 15.x=-9/14 17.x=20/3 19.x=-4/3 21.x=15 23.x=490/9
25.x=12 27.x=11/70 C.
29. x=1 31.x=0 D.
33. Identity 35. Inconsistent 37. Identity
Linear Word Problems Pg. 43. A. 1. (x – 10) 3. 2(x +10) 5.
2[x + (x – 100)] = 4x – 200 7. 3x 9. 3x +8(x +4) 11. (50 – x) lbs of cashews,
cost (400 – 5x) 13. A gets $75, B gets $45 15. A gets $48, B gets $14,
C gets $58 17. A gets $19.20, B gets $57.60, C gets $43.20 19. $20
21. 12 gal. 23. $4,000 at 4% and $6,000 at 6% 25. $7,500 at
each 27. 8 lbs. of peanuts, 4 lbs of cashews 29. 5 lbs. of
peanuts and 10 lbs. of cashews
Literal Equations Pg. 45 A. 1. b=a-d+e 3. b= (e-d) / 2 5. b= (e-d)/(2+a)
7.y=5/(3x+6) 9. y=(5-3x)/6x 11.t=6w 13. t=6w+3 15.t=(aw+b)/3
17. t=6(w-a)+3 19.t=[c(w-d)+b]/a
1. x < 3 3. x < –8 5. x ≥ 3
7. x ≥ –8 9. 3 ≤ x < 6 11. –8 ≤ x < 1
13. No such x 15. –7 ≤ x ≤ –3 17. x < –2
19. x < –3 21. –4 ≤ x 23. x ≤ –8/5
25. – 2/5 ≤ x 27. x ≤ – 7/2
Inequalities
Pg. 48
3 –8 3
–8 63 1–8
–3–7 –2
–3 –4 –8/5
–2/5
–7/2

101
The Rectangular Coordinate System Pg. 51. 1. A(5, 0), B(–3 , 0), C(0, 4), D(0, –3),
E(4, 2), F(–5, 3), G(–4 , –5 )
3. 5. 3rd
quadrant
7. Any ordered pairs with (+#, –#)
Linear Equations and Lines Pg. 54 1. y = 4 3. x = 3 5. y = 5/2 7. x = 5 9. y = 8/3
11. y = 3/2 13. x = 4
23. 25. .

102
27. 29.
31. 33.
Systems of Linear Equations Pg. 57 A. 1. a. 3. c.
B. 5. (0, 3) 7. (2, 1) C.
9. –x 11. x 13. –9y D.
15. (1, 2) 17. (3, 0) 19. (3, –3/2) 21. (1, 1) 23. (–18, –25)
25.(–2, –1) 27.(34, 72)
E. 29. Dependent
Systems of Linear Equations II Pg. 60 1. (1, 2) 3. (3, 0) 5. (3, –3/2)
7. 9. 11.
Linear Word Problems II Pg. 63 1. A gets $74, B gets $45 3. A gets $86, B gets $34 5.
4 lb –cashews, 8 lb–peanuts 7. $11,250 at 4% , $20,0

103
500 at 6% 9. $5/lb–cashews, $2/lb–peanuts 11. $2/lb each 13. $5,000 at 4%, $8,000
at 5% 15. $3,000 at 4%, $9,000 at 5%
Exponents Pg. 65 A.1. 1/2 3. 1/8 5. 1/27 7. 1/64 9. 3/2 11. 1/18 13. 1/6 15. -5/12
B. 17. 2
x 19. 8
x
21. 2 7
x y
23. 4 139
4
x y 
25. 17
x 27. 5 7
x y
29. 5
x 31. 6 3
x y
33. 4 3
4x y
35. 4 1
2x y 
C39. 2 2
3 x
41. 5x
b 
43. 4x
e 
Scientific Notation Pg. 68 A. 1.2410 3.3.15 5.0.00000762 7.0.00000593
B. 9.5310 11.0.031 13.5890000000 15.0.0000000052
C. 17. 1
1.0 10 19. 2
2.0 10 21. 5
2.31 10 23. 1
0.1 10 25. 2
2.0 10
 27. 6
2.31 10

D. 31. 1
0.1 10 1.0  33. 1
1.0 10 0.1
  35. 6 5 11
4 10 1.2 10 4.8 10 0.000000000048  
     
37. 3
4.8 10 4800  39. 4 5 2
2.4 10 6 10 1.44 10 144
     
41. 8 8 17
2.4 10 6 10 1.44 10 1440000000000000     
43.
4 5
9
4 4
2.4 10 6 10
1.2 10 1200000000
1.5 10 8 10

 
  
  
  
45.
4 5
13
2 2
2.4 10 5 10
10 0.0000000000001
1.5 10 8 10
 
  
 
  
Polynomial Expressions Pg. 71 A 1. 2 and -2 3. 2 and 2 5. 40 and -40 7.
80 and 80
B 9. 1 and 7 11. 22 and 28 13. -16 and 16 15.-5x-8 17.-33x+28 19.-5x-61
21. 2
11 7x x   23. 2
3 26 20x x   25. 2
26 16 25x x   27. 3 2
12 74 103 10x x x  
29. 2xy, when x = –1 we get –2, when x –1 and y = 2 we get 4. 31. 2
3 6x xy -9
Polynomial Expressions II Pg. 79 A. 1. 4
9x 3. 5
15x 5. 3
15x 7. 9
6x
B. 9. 8 5 4
x x x 
C. 11. 2
12 74 63x x  13. 2
65 34x x  15. 2
26 7 6x x  
D. 19. 2
2 35x x  21. 2
12 35x x  23. 2
3 7 40x x   25. 2
9 49x 
27. 3 2
3 8 56 14x x x    29. 3 2
10 22 15x x x   31. 2
2 1x x  33. 2
25 40 16x x 
35. 3 2
24 20 24x x x  37. 3 2
6 7 9 2x x x   39. 3
1x  41. 5
1x 
Special Binomial Operations Pg. 74 A. 1. 2
12 35x x  3. 2
2 35x x  5. 2
6 2 20x x 
7. 2
4 25x  9. 2
12 19 21x x   11. 2
4 25x  13. 2
45 2 8x x  15. 2
20 19 3x x 
17. 2 2
2 35x xy y  19. 2 2
9 49x y
B. 21. 2
2 15x x   23. 2
16 2 3x x  

104
C. 25. 2
5 12 35x x  27. 2
7 17 34x x  29. 2
7 36 5x x  31. 2
2 31 5x x  
33. 2
2 13 1x x  35. 2
3 26 40x x  
Multiplication Formulas Pg. 77 A. 1. 399 3. 1599 5. 396 7. 2496
B. 9. 441 11. 1521 13. 9604 15. 10100.25
C. 19. 2
49x  21. 2
9 25x  23. 2
49 9x 25. 2 2
4 9x y 27. 2
1 49y 29. 2
100 81x 31.
2
14 49x x  33. 2 2
9 30 25x xy y  35. 2 2
4 4x xh h 
Polynomials Formulas II Pg. 83 See last section answers.
Factoring Out the GCF Pg. 80 A.1. 2 3. 4 5. 2x 7. 2 2
4A B
B. 9. 2(2 3 )y 11. 4(8 5 3 )A B C  13. 2 (2 3 )x x  15. 2 2
4 (8 5 3 )A B B AB C 
17. 2 2
2 (2 4 1)x x x  .19. ( 2)( 3)x x 
C.21. 2( 2)y  23. 5( 3)x  D.
25.( 3)( 2)y y  27.( 2)( 3)y y  29.( 4)( 1)y y  31. ( 3)(2 1)y y  33.(2 3)(2 3)x x  35.
(2 )(5 3)y y  37.( 3 )(2 )y x y x  39.( 4 )( 5 2 )x y x y  
Factoring Trinomials I Pg. 82 A. 1.  1 2x x  3.  2 3x x  5. 2
2x x  is prime
7.   4 2x x  9.   2 3x x  11.   6 1x x  13.   4 5x x  15. Prime
17.  2 12x x  19.  3 8x x  21. Prime 23. Prime 25.  6 6x x 
B. 27.   2 7x x   29.   3 2 12x x  31.   2
2 3 3x x x   33.   3
5 1 2 1x x x 
C. 35.   3 2 8x x x   37.   12x y x y   39.   12 3x y x y   D.
41. 2
1x  is prime 43. 2
9x  is prime 45. 2
4 25x  is prime
Factoring Trinomials II Pg. 85 A. 1. ( 1)(3 2)x x  3.( 1)(3 1)x x  5.( 1)(2 1)x x 
7.( 2)(2 1)x x  9. ( 1)(5 2)x x  11.( 1)(3 2)x x  13( 1)(3 2)x x  15.
(2 3)(3 2)x x  17.( 2)(6 1)x x  19. (2 1)(3 2)x x  21.(2 3)(3 2)x x 
23.(2 1)(3 4)x x  125 .(3 8)(2 1)x x  225 .(2 3)(2 3)x x  27.(5 2)(5 2)x x 
29. prime B. 31. ( 2)(2 1)x x x  33. 10 (4 1)x x 
Factoring Trinomials II Pg. 87 A. 1– 27, odd are factorable, 29 is prime B.
31. .Not factorable 33. –5(3x + 2)(x + 1)
Substitution and Factoring Formulas Pg. 90 A. 1.  1 1x x  3.  2 1 2 1xy xy  5.
3(3 2 )x 7. 4( 2)x  9.   2
2 2 4x x x   11.   2 2
2 3 4 6 9x y x xy y  
13.   2
4 5 16 20 25x x x   15. 2
2 10h h 
B. 17.   1 1x x  19.   3 3x x  21.   5 8 5 8x x  23.   2 1 2 1xy xy 
25.   2 9 2 9xy xy  27.   2 5 3 5 3x y x y  29. 2z(x+7)(x–7)

105
C. 31.   2
1 1x x x   33.   2
3 9 3y y y   35.   2
4 16 4y y y  
37.  2
5 5 25x x x   39.  2
10 10 100x x x  
Applications of Factoring Pg . 92 A.1.6,-4,6 3.
7 1
,7,
4 4
 5.20,-18,42
B. 7.+, – , – , + 9. – , + , – , + 11. +, + ,+ , –
C. 13. x =–1, 4 15. x = – 3, – 4 17. x = 3, – 3 19. x = 5, – 2 21 .x = – a, 2
23. x = +1/2, –1/2 25. x = 2, 2 27.x = –1, – 5 29. x = 0, 4, –2
31. x = 0, 0, 4 33. x =0, 0, 2, 2 35. x = ½, –3 37. x = 0, 4 39. x = 2, – 2
Word Problems Pg . 96 A. 1. After 1 second and 3 seconds 3. a. After 1 second and
5 seconds b. After 3 seconds c. 144 ft B. 5. x = 2 ft 7. x = 3 ft
9. x = 3 ft 11. x = 2 ft 13. x = 4ft 15.x = 3 ft

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