Factoring Polynomials using the difference of two squares2.ppt

Martin-Gay, Developmental Mathematics 2
13.1 – The Greatest Common Factor
13.2 – Factoring Trinomials of the Form x2
+ bx + c
13.3 – Factoring Trinomials of the Form ax2
+ bx + c
13.4 – Factoring Trinomials of the Form x2
+ bx + c
by Grouping
13.5 – Factoring Perfect Square Trinomials and
Difference of Two Squares
13.6 – Solving Quadratic Equations by Factoring
13.7 – Quadratic Equations and Problem Solving
Chapter Sections

§ 13.1
The Greatest Common
Factor

Factors
Factors (either numbers or polynomials)
When an integer is written as a product of
integers, each of the integers in the product is a
factor of the original number.
When a polynomial is written as a product of
polynomials, each of the polynomials in the
product is a factor of the original polynomial.
Factoring – writing a polynomial as a product of
polynomials.

Greatest common factor – largest quantity that is a
factor of all the integers or polynomials involved.
Finding the GCF of a List of Integers or Terms
1) Prime factor the numbers.
2) Identify common prime factors.
3) Take the product of all common prime factors.
• If there are no common prime factors, GCF is 1.
Greatest Common Factor

Find the GCF of each list of numbers.
1) 12 and 8
12 = 2 · 2 · 3
8 = 2 · 2 · 2
So the GCF is 2 · 2 = 4.
2) 7 and 20
7 = 1 · 7
20 = 2 · 2 · 5
There are no common prime factors so the
GCF is 1.
Example

Find the GCF of each list of numbers.
1) 6, 8 and 46
6 = 2 · 3
8 = 2 · 2 · 2
46 = 2 · 23
So the GCF is 2.
2) 144, 256 and 300
144 = 2 · 2 · 2 · 3 · 3
256 = 2 · 2 · 2 · 2 · 2 · 2 · 2 · 2
300 = 2 · 2 · 3 · 5 · 5
So the GCF is 2 · 2 = 4.
Example

1) x3
and x7
x3
= x · x · x
x7
= x · x · x · x · x · x · x
So the GCF is x · x · x = x3
t 6x5
and 4x3
6x5
= 2 · 3 · x · x · x
4x3
= 2 · 2 · x · x · x
So the GCF is 2 · x · x · x = 2x3
Find the GCF of each list of terms.
Example

Find the GCF of the following list of terms.
a3
b2
, a2
b5
and a4
b7
a3
b2
= a · a · a · b · b
a2
b5
= a · a · b · b · b · b · b
a4
b7
= a · a · a · a · b · b · b · b · b · b · b
So the GCF is a · a · b · b = a2
b2
Notice that the GCF of terms containing variables will use the
smallest exponent found amongst the individual terms for each
variable.
Example

The first step in factoring a polynomial is to
find the GCF of all its terms.
Then we write the polynomial as a product by
factoring out the GCF from all the terms.
The remaining factors in each term will form a
polynomial.
Factoring Polynomials

Factor out the GCF in each of the following
polynomials.
1) 6x3
– 9x2
+ 12x =
3 · x · 2 · x2
– 3 · x · 3 · x + 3 · x · 4 =
3x(2x2
– 3x + 4)
2) 14x3
y + 7x2
y – 7xy =
7 · x · y · 2 · x2
+ 7 · x · y · x – 7 · x · y · 1 =
7xy(2x2
+ x – 1)
Factoring out the GCF
Example

Factor out the GCF in each of the following
polynomials.
1) 6(x + 2) – y(x + 2) =
6 · (x + 2) – y · (x + 2) =
(x + 2)(6 – y)
2) xy(y + 1) – (y + 1) =
xy · (y + 1) – 1 · (y + 1) =
(y + 1)(xy – 1)
Factoring out the GCF
Example

Remember that factoring out the GCF from the terms of
a polynomial should always be the first step in factoring
a polynomial.
This will usually be followed by additional steps in the
process.
Factor 90 + 15y2
– 18x – 3xy2
.
90 + 15y2
– 18x – 3xy2
= 3(30 + 5y2
– 6x – xy2
) =
3(5 · 6 + 5 · y2
– 6 · x – x · y2
) =
3(5(6 + y2
) – x (6 + y2
)) =
3(6 + y2
)(5 – x)
Factoring
Example

§ 13.2
Factoring Trinomials of the
Form x2
+ bx + c

Factoring Trinomials
Recall by using the FOIL method that
F O I L
(x + 2)(x + 4) = x2
+ 4x + 2x + 8
= x2
+ 6x + 8
To factor x2
+ bx + c into (x + one #)(x + another #),
note that b is the sum of the two numbers and c is the
product of the two numbers.
So we’ll be looking for 2 numbers whose product is c
and whose sum is b.
Note: there are fewer choices for the product, so that’s
why we start there first.

Factor the polynomial x2
+ 13x + 30.
Since our two numbers must have a product of 30 and a
sum of 13, the two numbers must both be positive.
Positive factors of 30 Sum of Factors
1, 30 31
2, 15 17
3, 10 13
Note, there are other factors, but once we find a pair
that works, we do not have to continue searching.
So x2
+ 13x + 30 = (x + 3)(x + 10).
Example

– 11x + 24.
sum of -11, the two numbers must both be negative.
Negative factors of 24 Sum of Factors
– 1, – 24 – 25
– 2, – 12 – 14
– 3, – 8 – 11
So x2
– 11x + 24 = (x – 3)(x – 8).
Example

– 2x – 35.
Since our two numbers must have a product of – 35 and a
sum of – 2, the two numbers will have to have different signs.
Factors of – 35 Sum of Factors
– 1, 35 34
1, – 35 – 34
– 5, 7 2
5, – 7 – 2
So x2
– 2x – 35 = (x + 5)(x – 7).
Example

– 6x + 10.
sum of – 6, the two numbers will have to both be negative.
Negative factors of 10 Sum of Factors
– 1, – 10 – 11
– 2, – 5 – 7
Since there is not a factor pair whose sum is – 6,
x2
– 6x +10 is not factorable and we call it a prime
polynomial.
Prime Polynomials
Example

You should always check your factoring
results by multiplying the factored polynomial
to verify that it is equal to the original
polynomial.
Many times you can detect computational
errors or errors in the signs of your numbers
by checking your results.
Check Your Result!

§ 13.3
Factoring Trinomials of
the Form ax2
+ bx + c

Factoring Trinomials
Returning to the FOIL method,
F O I L
(3x + 2)(x + 4) = 3x2
+ 12x + 2x + 8
= 3x2
+ 14x + 8
To factor ax2
+ bx + c into (#1·x + #2)(#3·x + #4), note
that a is the product of the two first coefficients, c is
the product of the two last coefficients and b is the
sum of the products of the outside coefficients and
inside coefficients.
Note that b is the sum of 2 products, not just 2
numbers, as in the last section.

Factor the polynomial 25x2
+ 20x + 4.
Possible factors of 25x2
are {x, 25x} or {5x, 5x}.
Possible factors of 4 are {1, 4} or {2, 2}.
We need to methodically try each pair of factors until we find
a combination that works, or exhaust all of our possible pairs
of factors.
Keep in mind that, because some of our pairs are not identical
factors, we may have to exchange some pairs of factors and
make 2 attempts before we can definitely decide a particular
pair of factors will not work.
Example
Continued.

We will be looking for a combination that gives the sum of the
products of the outside terms and the inside terms equal to 20x.
{x, 25x} {1, 4} (x + 1)(25x + 4) 4x 25x 29x
(x + 4)(25x + 1) x 100x 101x
{x, 25x} {2, 2} (x + 2)(25x + 2) 2x 50x 52x
Factors
of 25x2
Resulting
Binomials
Product of
Outside Terms
Product of
Inside Terms
Sum of
Products
Factors
of 4
{5x, 5x} {2, 2} (5x + 2)(5x + 2) 10x 10x 20x
Example Continued
Continued.

Check the resulting factorization using the FOIL method.
(5x + 2)(5x + 2) =
= 25x2
+ 10x + 10x + 4
5x(5x)
F
+ 5x(2)
O
+ 2(5x)
I
+ 2(2)
L
= 25x2
+ 20x + 4
So our final answer when asked to factor 25x2
+ 20x + 4
will be (5x + 2)(5x + 2) or (5x + 2)2
.
Example Continued

– 41x + 10.
Possible factors of 21x2
are {x, 21x} or {3x, 7x}.
Since the middle term is negative, possible factors of
10 must both be negative: {-1, -10} or {-2, -5}.
We need to methodically try each pair of factors until
we find a combination that works, or exhaust all of our
possible pairs of factors.
Example
Continued.

We will be looking for a combination that gives the sum of
the products of the outside terms and the inside terms equal
to 41x.
Factors
of 21x2
Resulting
Binomials
Product of
Outside Terms
Product of
Inside Terms
Sum of
Products
Factors
of 10
{x, 21x}{1, 10}(x – 1)(21x – 10) –10x 21x – 31x
(x – 10)(21x – 1) –x 210x – 211x
{x, 21x} {2, 5} (x – 2)(21x – 5) –5x 42x – 47x
(x – 5)(21x – 2) –2x 105x – 107x
Example Continued
Continued.

Factors
of 21x2
Resulting
Binomials
Product of
Outside Terms
Product of
Inside Terms
Sum of
Products
Factors
of 10
(3x – 5)(7x – 2) 6x 35x 41x
{3x, 7x}{1, 10}(3x – 1)(7x – 10) 30x 7x 37x
(3x – 10)(7x – 1) 3x 70x 73x
{3x, 7x} {2, 5} (3x – 2)(7x – 5) 15x 14x 29x
Example Continued
Continued.

(3x – 5)(7x – 2) =
= 21x2
– 6x – 35x + 10
3x(7x)
F
+ 3x(-2)
O
- 5(7x)
I
- 5(-2)
L
= 21x2
– 41x + 10
So our final answer when asked to factor 21x2
– 41x + 10
will be (3x – 5)(7x – 2).
Example Continued

– 7x + 6.
The only possible factors for 3 are 1 and 3, so we know that, if
factorable, the polynomial will have to look like (3x )(x )
in factored form, so that the product of the first two terms in the
binomials will be 3x2
.
Since the middle term is negative, possible factors of 6 must
both be negative: {1,  6} or { 2,  3}.
We need to methodically try each pair of factors until we find a
combination that works, or exhaust all of our possible pairs of
factors.
Example
Continued.

We will be looking for a combination that gives the sum of the
products of the outside terms and the inside terms equal to 7x.
{1, 6} (3x – 1)(x – 6) 18x x 19x
(3x – 6)(x – 1) Common factor so no need to test.
{2, 3} (3x – 2)(x – 3) 9x 2x 11x
(3x – 3)(x – 2) Common factor so no need to test.
Factors
of 6
Resulting
Binomials
Product of
Outside Terms
Product of
Inside Terms
Sum of
Products
Example Continued
Continued.

Now we have a problem, because we have
exhausted all possible choices for the factors,
but have not found a pair where the sum of the
products of the outside terms and the inside
terms is –7.
So 3x2
– 7x + 6 is a prime polynomial and will
not factor.
Example Continued

y2
– 2xy2
– 60y2
.
Remember that the larger the coefficient, the greater the
probability of having multiple pairs of factors to check.
So it is important that you attempt to factor out any
common factors first.
6x2
y2
– 2xy2
– 60y2
= 2y2
(3x2
– x – 30)
The only possible factors for 3 are 1 and 3, so we know
that, if we can factor the polynomial further, it will have to
look like 2y2
(3x )(x ) in factored form.
Example
Continued.

Since the product of the last two terms of the binomials
will have to be –30, we know that they must be
different signs.
Possible factors of –30 are {–1, 30}, {1, –30}, {–2, 15},
{2, –15}, {–3, 10}, {3, –10}, {–5, 6} or {5, –6}.
We will be looking for a combination that gives the sum
of the products of the outside terms and the inside terms
equal to –x.
Example Continued
Continued.

Factors
of -30
Resulting
Binomials
Product of
Outside Terms
Product of
Inside Terms
Sum of
Products
{-1, 30} (3x – 1)(x + 30) 90x -x 89x
(3x + 30)(x – 1) Common factor so no need to test.
{1, -30} (3x + 1)(x – 30) -90x x -89x
(3x – 30)(x + 1) Common factor so no need to test.
{-2, 15} (3x – 2)(x + 15) 45x -2x 43x
(3x + 15)(x – 2) Common factor so no need to test.
{2, -15} (3x + 2)(x – 15) -45x 2x -43x
(3x – 15)(x + 2) Common factor so no need to test.
Example Continued
Continued.

Factors
of –30
Resulting
Binomials
Product of
Outside Terms
Product of
Inside Terms
Sum of
Products
{–3, 10} (3x – 3)(x + 10) Common factor so no need to test.
(3x + 10)(x – 3) –9x 10x x
{3, –10} (3x + 3)(x – 10) Common factor so no need to test.
(3x – 10)(x + 3) 9x –10x –x
Example Continued
Continued.

(3x – 10)(x + 3) =
= 3x2
+ 9x – 10x – 30
3x(x)
F
+ 3x(3)
O
– 10(x)
I
– 10(3)
L
= 3x2
– x – 30
So our final answer when asked to factor the polynomial
6x2
y2
– 2xy2
– 60y2
will be 2y2
(3x – 10)(x + 3).
Example Continued

§ 13.4
Factoring Trinomials of
the Form x2
+ bx + c
by Grouping

Factoring polynomials often involves additional
techniques after initially factoring out the GCF.
One technique is factoring by grouping.
Factor xy + y + 2x + 2 by grouping.
Notice that, although 1 is the GCF for all four
terms of the polynomial, the first 2 terms have a
GCF of y and the last 2 terms have a GCF of 2.
xy + y + 2x + 2 = x · y + 1 · y + 2 · x + 2 · 1 =
y(x + 1) + 2(x + 1) = (x + 1)(y + 2)
Factoring by Grouping
Example

Factoring a Four-Term Polynomial by Grouping
1) Arrange the terms so that the first two terms have a
common factor and the last two terms have a common
factor.
2) For each pair of terms, use the distributive property to
factor out the pair’s greatest common factor.
3) If there is now a common binomial factor, factor it out.
4) If there is no common binomial factor in step 3, begin
again, rearranging the terms differently.
• If no rearrangement leads to a common binomial
factor, the polynomial cannot be factored.

1) x3
+ 4x + x2
+ 4 = x · x2
+ x · 4 + 1 · x2
+ 1 · 4 =
x(x2
+ 4) + 1(x2
+ 4) =
(x2
+ 4)(x + 1)
2) 2x3
– x2
– 10x + 5 = x2
· 2x – x2
· 1 – 5 · 2x – 5 · (– 1) =
x2
(2x – 1) – 5(2x – 1) =
(2x – 1)(x2
– 5)
Factor each of the following polynomials by grouping.
Example

Factor 2x – 9y + 18 – xy by grouping.
Neither pair has a common factor (other than 1).
So, rearrange the order of the factors.
2x + 18 – 9y – xy = 2 · x + 2 · 9 – 9 · y – x · y =
2(x + 9) – y(9 + x) =
2(x + 9) – y(x + 9) = (make sure the factors are identical)
(x + 9)(2 – y)
Example

§ 13.5
Factoring Perfect Square
Trinomials and the

Recall that in our very first example in Section
4.3 we attempted to factor the polynomial
25x2
+ 20x + 4.
The result was (5x + 2)2
, an example of a
binomial squared.
Any trinomial that factors into a single
binomial squared is called a perfect square
trinomial.
Perfect Square Trinomials

In the last chapter we learned a shortcut for squaring a
binomial
(a + b)2
= a2
+ 2ab + b2
(a – b)2
= a2
– 2ab + b2
So if the first and last terms of our polynomial to be
factored are can be written as expressions squared, and
the middle term of our polynomial is twice the product
of those two expressions, then we can use these two
previous equations to easily factor the polynomial.
a2
+ 2ab + b2
=(a + b)2
a2
– 2ab + b2
= (a – b)2

– 8xy + y2
.
Since the first term, 16x2
, can be written as (4x)2
, and
the last term, y2
is obviously a square, we check the
middle term.
8xy = 2(4x)(y) (twice the product of the expressions
that are squared to get the first and last terms of the
polynomial)
Therefore 16x2
– 8xy + y2
= (4x – y)2
.
Note: You can use FOIL method to verify that the
factorization for the polynomial is accurate.
Example

Another shortcut for factoring a trinomial is when we
want to factor the difference of two squares.
a2
– b2
= (a + b)(a – b)
A binomial is the difference of two square if
1.both terms are squares and
2.the signs of the terms are different.
9x2
– 25y2
– c4
+ d4

Example
– 9.
The first term is a square and the last term, 9, can be
written as 32
. The signs of each term are different, so
we have the difference of two squares
Therefore x2
– 9 = (x – 3)(x + 3).
Note: You can use FOIL method to verify that the
factorization for the polynomial is accurate.

§ 13.6
Solving Quadratic
Equations by Factoring

Zero Factor Theorem
Quadratic Equations
• Can be written in the form ax2
+ bx + c = 0.
• a, b and c are real numbers and a  0.
• This is referred to as standard form.
Zero Factor Theorem
• If a and b are real numbers and ab = 0, then a = 0
or b = 0.
• This theorem is very useful in solving quadratic
equations.

Steps for Solving a Quadratic Equation by
Factoring
1) Write the equation in standard form.
2) Factor the quadratic completely.
3) Set each factor containing a variable equal to 0.
4) Solve the resulting equations.
5) Check each solution in the original equation.
Solving Quadratic Equations

Solve x2
– 5x = 24.
• First write the quadratic equation in standard form.
x2
– 5x – 24 = 0
• Now we factor the quadratic using techniques from
the previous sections.
x2
– 5x – 24 = (x – 8)(x + 3) = 0
• We set each factor equal to 0.
x – 8 = 0 or x + 3 = 0, which will simplify to
x = 8 or x = – 3
Example
Continued.

• Check both possible answers in the original
equation.
82
– 5(8) = 64 – 40 = 24 true
(–3)2
– 5(–3) = 9 – (–15) = 24 true
• So our solutions for x are 8 or –3.
Example Continued

Solve 4x(8x + 9) = 5
• First write the quadratic equation in standard form.
32x2
+ 36x = 5
32x2
+ 36x – 5 = 0
• Now we factor the quadratic using techniques from the
previous sections.
32x2
+ 36x – 5 = (8x – 1)(4x + 5) = 0
• We set each factor equal to 0.
8x – 1 = 0 or 4x + 5 = 0
Example
Continued.
8x = 1 or 4x = – 5, which simplifies to x = or
5
.
4

1
8

• Check both possible answers in the original equation.
   
      
1 1 1
4 8 9 4 1 9 4 (10) (10) 5
8
1
8
1
8 8 2
     
true
   
      
5 5
4 8 9 4 10 9 4 ( 1) ( 5)( 1) 5
4
5 5
4 4
4
          
 

true
• So our solutions for x are or .
8
1
4
5

Example Continued

Recall that in Chapter 3, we found the x-intercept of linear
equations by letting y = 0 and solving for x.
The same method works for x-intercepts in quadratic
equations.
Note: When the quadratic equation is written in standard
form, the graph is a parabola opening up (when a > 0) or
down (when a < 0), where a is the coefficient of the x2
term.
The intercepts will be where the parabola crosses the x-
axis.
Finding x-intercepts

Find the x-intercepts of the graph of y = 4x2
+ 11x + 6.
The equation is already written in standard form, so
we let y = 0, then factor the quadratic in x.
0 = 4x2
+ 11x + 6 = (4x + 3)(x + 2)
We set each factor equal to 0 and solve for x.
4x + 3 = 0 or x + 2 = 0
4x = –3 or x = –2
x = –¾ or x = –2
So the x-intercepts are the points (–¾, 0) and (–2, 0).
Finding x-intercepts
Example

§ 13.7
Quadratic Equations
and Problem Solving

Strategy for Problem Solving
General Strategy for Problem Solving
1) Understand the problem
• Read and reread the problem
• Choose a variable to represent the unknown
• Construct a drawing, whenever possible
• Propose a solution and check
2) Translate the problem into an equation
3) Solve the equation
4) Interpret the result
• Check proposed solution in problem
• State your conclusion

The product of two consecutive positive integers is 132. Find the
two integers.
1.) Understand
Read and reread the problem. If we let
x = one of the unknown positive integers, then
x + 1 = the next consecutive positive integer.
Finding an Unknown Number
Example
Continued

Example continued
2.) Translate
Continued
two consecutive positive integers
x (x + 1)
is
=
132
132
•
The product of

Example continued
3.) Solve
Continued
x(x + 1) = 132
x2
+ x = 132 (Distributive property)
x2
+ x – 132 = 0 (Write quadratic in standard form)
(x + 12)(x – 11) = 0 (Factor quadratic polynomial)
x + 12 = 0 or x – 11 = 0 (Set factors equal to 0)
x = –12 or x = 11 (Solve each factor for x)

Example continued
4.) Interpret
Check: Remember that x is suppose to represent a positive
integer. So, although x = -12 satisfies our equation, it cannot be a
solution for the problem we were presented.
If we let x = 11, then x + 1 = 12. The product of the two numbers
is 11 · 12 = 132, our desired result.
State: The two positive integers are 11 and 12.

Pythagorean Theorem
In a right triangle, the sum of the squares of the
lengths of the two legs is equal to the square of the
length of the hypotenuse.
(leg a)2
+ (leg b)2
= (hypotenuse)2
leg a
hypotenuse
leg b
The Pythagorean Theorem

Find the length of the shorter leg of a right triangle if the longer leg
is 10 miles more than the shorter leg and the hypotenuse is 10 miles
less than twice the shorter leg.
Example
Continued
1.) Understand
Read and reread the problem. If we let
x = the length of the shorter leg, then
x + 10 = the length of the longer leg and
2x – 10 = the length of the hypotenuse.
x
+ 10
2 - 10
x
x

Example continued
2.) Translate
Continued
By the Pythagorean Theorem,
(leg a)2
+ (leg b)2
= (hypotenuse)2
x2
+ (x + 10)2
= (2x – 10)2
3.) Solve
x2
+ (x + 10)2
= (2x – 10)2
x2
+ x2
+ 20x + 100 = 4x2
– 40x + 100 (multiply the binomials)
2x2
+ 20x + 100 = 4x2
– 40x + 100 (simplify left side)
x = 0 or x = 30 (set each factor = 0 and solve)
0 = 2x(x – 30) (factor right side)
0 = 2x2
– 60x (subtract 2x2
+ 20x + 100 from both sides)

Example continued
4.) Interpret
Check: Remember that x is suppose to represent the length of
the shorter side. So, although x = 0 satisfies our equation, it
cannot be a solution for the problem we were presented.
If we let x = 30, then x + 10 = 40 and 2x – 10 = 50. Since 302 +
402 = 900 + 1600 = 2500 = 502, the Pythagorean Theorem
checks out.
State: The length of the shorter leg is 30 miles. (Remember that
is all we were asked for in this problem.)

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