Factoring 15.3 and 15.4 Grouping and Trial and Error

15.3 Factoring15.3 Factoring
Trinomials (Part 2)Trinomials (Part 2)
Ax2
+ bx + c

METHODS
1.Trial and Error (Takes the most time)
2.Factoring by Grouping
3.Box Method (Grouping with a Box)

Lehmann, Elementary and Intermediate Algebra, 1edSection 8.3 Slide 10
Copyright © 2011 Pearson Education, Inc.
Method 1: Factoring Trinomials by Trial and Error
Factoring by Grouping
Factor
Example
2
3 14 8.x x+ +

Factor 3x2
+ 14x + 8, the result will be of the form
(3x +?)(x +?)
Product of the last terms must be 8, so the last terms
must be 1 and 8 or 2 and 4
Rule out negative last terms in the factors, because the
middle term of 3x
2
+14x +8 has the positive coefficient
14
Solution

• Decide between the two pairs of possible last terms
by multiplying:
Solution Continued

Method 1: Factoring Trinomials by Trial and ErrorMethod 1: Factoring Trinomials by Trial and Error
Factoring by Trial and ErrorFactoring by Trial and Error
FactorFactor
ExampleExample
2
2 5 25.x x− −

Factoring by Trial and Error
Factor 2x
2
– 5 – 25, the result will be of the form
(2x +?)(x +?)
Product of the last terms must be –25, so the last terms
must be 1 and –25, 5 and –5, or –1 and 25
Decide amongst the three pairs of possible last
terms by multiplying:
Solution

Factoring by Trial and Error
Solution Continued

Factoring Trinomials Review
 X2
+ 6x + 5
 (x + )(x + )
 Find factors of 5 that add to 6:
1*5 = 5 1+5 = 6
 (x + 1)(x + 5)

Factoring Trinomials where a ≠ 1
BY GROUPING! IT’S FASTER!
 Follow these steps:
1. Find two numbers that multiply to ac and add to
b for ax2
+ bx + c
2. Replace bx with the sum of the 2 factors found
in step 1.
ie: ax2
+ bx + c becomes ax2
+ mx + nx + c, where m
and n are the factors found in step 1.
3. Use grouping to factor this expression into 2
binomials

2x2
+ 5x + 2
Step 1: ac = 2*2 = 4
1*4 = 4 1+4 = 5
2*2 = 4 2+2 = 4
m = 1 and n = 4
Step 2: Rewrite our trinomial by expanding bx
2x2
+ 1x + 4x + 2
Step 3: Group and Factor
(2x2
+ 1x) + (4x + 2)
x(2x + 1) + 2( 2x + 1)
(2x + 1) (x + 2)

2x2
+ 5x + 2
Questions for thought:
1. Does it matter which order the new factors are
entered into the polynomial?
2. Do the parenthesis still need to be the same?
3. Will signs continue to matter when finding m
and n?
4. Does it matter how we group the terms for
factoring?

3z2
+ z – 2
Step 1: ac = 3*-2 = -6
-1*6 = -6 -1+6 = 5 1* -6 = -6 1+-6 = -5
-2*3 = -6 -2+3 = 1 2* -3 = -6 2+-3 = -1
m = -2 and n = 3
3z2
+ 3z – 2z – 2
Step 3: Group and Factor
(3z2
+ 3z) + (-2z - 2)
3z(z + 1) - 2( z + 1)
(z + 1) (3z - 2)

3z2
+ z – 2
Step 1: ac = 3*2 = 6
-1*6 = -6 -1+7 = 6 1* -6 = -6 1+-7 = -6
-2*3 = -6 -2+3 = 1 2* -3 = -6 2+-3 = -1
m = -2 and n = 3
3z2
+ 3z – 2z – 2
Notice that I changed the order of m and n between
step 1 and step 2. Why do you think I did this? Do
you have to change the order to get the correct
answer?

3z2
+ z – 2
 What are the 3 steps for factoring this quadratic
equation?
 Step 1: Multiply a*c. Find the factors that multiply
to ac and add to b
 Step 2: Expand bx to equal mx + nx
 Step 3: Group and Factor

4x3
– 22x2
+ 30x
 Step 0: Factor out the GCF: 2x
2x(2x2
– 11x + 15)
 Step 1: a*c = 30
-1*-30 = 30 -1+-30 = -31
-2*-15 = 30 -2+-15 = -17
-3*-10 = 30 -3+-10 = -13
-5*-6 = 30 -5+-6 = -11

4x3
– 22x2
+ 30x
-11x = -5x + -6x
2x(2x2
– 5x – 6x + 15)
2x((2x2
– 5x )(– 6x + 15))
2x(x(2x – 5) -3(2x – 5))
2x(2x – 5) (x – 3)

4x3
– 22x2
+ 30x
-11x = -5x + -6x
2x(2x2
– 5x – 6x + 15)
2x((2x2
– 5x )(– 6x + 15))
2x(x(2x – 5) -3(2x – 5))
Note: The Parenthesis are the Same
2x(2x – 5) (x – 3)

Practice
1. 3x2
+ 5x + 2
2. 6x2
+ 7x – 3
3. 6 + 4y2
– 11y

Practice
1. 3x2
+ 5x + 2
(3x + 2)(x + 1)
2. 6x2
+ 7x – 3
(3x – 1)(2x + 3)
3. 6 + 4y2
– 11y
(4y – 3)(y – 2)

Review
 What is Step 0? When do you need to include
this step?
 When will your factors both be negative?
 When will you have one negative and one
positive factor?
 How do you check your answers?

When the leading coefficient is negative,
factor out –1 from each term before using
other factoring methods.

When you factor out –1 in an early step, you
must carry it through the rest of the steps and
into the answer.
Caution!

Additional Example 4: Factoring ax2
+ bx + c
When a is Negative
Factor –2x2
– 5x – 3.
–1(2x2
+ 5x + 3)
–1( x + )( x+ )
Factor out –1.
a = 2 and c = 3;
Outer + Inner = 5
Factors of 2 Factors of 3 Outer + Inner
1 and 2 3 and 1 1(1) + 3(2) = 7
1 and 2 1 and 3 1(3) + 1(2) = 5
–1(x + 1)(2x + 3)
(x + 1)(2x + 3)

Additional Example 4: Factoring ax2
+ bx + c When a is
Negative
Factor –2x2
– 5x – 3.
–1(2x2
+ 5x + 3)
–1( x + )( x+ )
Factor out –1.
a = 2 and c = 3; Outer + Inner = 5
Factors of 2 Factors of 3 Outer + Inner
1 and 2 3 and 1 1(1) + 3(2) = 7
1 and 2 1 and 3 1(3) + 1(2) = 5
–1(x + 1)(2x + 3)
(x + 1)(2x + 3)

15.4 Special Types of
Factoring

1. Difference of Squares
2. Perfect Square Trinomials

Difference of Squares
 Think back to Chapter 5. What happened when
we multiplied a sum and difference?
(a – b)(a + b) = a2
– b2
 So, the reverse is also true.
a2
– b2
= (a – b)(a + b)

x2
– 25
 Notice that we do not have a bx term. This
means that we only have the F and L in foil;
therefore, none of the procedures from 6.1, 6.2,
or 6.3 will work.
 We need to use a2
– b2
= (a – b)(a + b)
where a = x and b = 5
 X2
– 25 = (x – 5)(x + 5)

x2
– 36
 We need to use a2
– b2
= (a – b)(a + b)
where a = x and b = 6
 X2
– 36 = (x – 6)(x + 6)

Practice
 4x2
– 9
 100 – 16t2
 49y2
– 64z2

Practice
 4x2
– 9
a = 2x, b = 3
(2x – 3) (2x + 3)
 100 – 16t2
a = 10, b = 4t
(10 – 4t) (10 + 4t)
 49y2
– 64z2
a = 7y, b = 8z
(7y – 8z) (7y + 8z)

Perfect Square Trinomials
 Think back to Chapter 5. What happened when
we squared a binomial?
(a + b)2
= a2
+ 2ab + b2
(a – b)2
= a2
– 2ab + b2
 So, the reverse is also true.
a2
+ 2ab + b2
= (a + b)2
a2
– 2ab + b2
= (a – b)2

x2
+ 10x + 25
 This can be worked 2 different ways
 The first way is the simplest, but depends on
whether you recognize the equation as a perfect
square trinomial.
a2
+ 2ab + b2
= (a + b)2
Where a = x and b = 5
x2
+ 10x + 25 = (x + 5)2

x2
+ 10x + 25
 The second way is to use the method we learned in
6.2
x2
+ 10x + 25
5*5 = 25 and 5+5 = 10
(x + 5) (x + 5) or (x + 5)2

4x2
- 4x + 1
 The first way is the simplest, but depends on
whether you recognize the equation as a perfect
square trinomial.
a2
+ 2ab + b2
= (a + b)2
Where a = 2x and b = 1
4x2
- 4x + 1 = (2x – 1)2

4x2
- 4x + 1
 This time we need to use the 6.3 method
4*1 = 4
-2 * -2 = 4 and -2 + -2 = -4
(4x2
– 2x) ( – 2x + 1)
2x(2x – 1) – 1(2x – 1)
(2x – 1) (2x – 1) or (2x – 1)2

Practice
 x2
– 4xy + 4y2
 9a2
– 60a + 100
 25y2
+ 20yz + 4z2

Practice
 x2
– 4xy + 4y2
a = x, b = 2y
(x – 2y)2
 9a2
– 60a + 100
a = 3a, b = 10
(3a – 10)
 25y2
+ 20yz + 4z2
a = 5y, b = 2z
(5y + 2z)

Review
 What methods can you use to factor a
Difference of Squares?
 What methods can you use to factor a Perfect
Square Trinomial?
 What clues should you look for to identify a
Difference of Squares?
 What clues should you look for to identify a
Perfect Square Trinomial?

Factoring 15.3 and 15.4 Grouping and Trial and Error

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Factoring 15.3 and 15.4 Grouping and Trial and Error