Exponents,
Polynomials and
Functions
Copyright © Cengage Learning. All rights reserved.
3
3.5
Special Factoring Techniques
Copyright © Cengage Learning. All rights reserved.
Objectives

Factor perfect square trinomials.

Factor difference of squares.

Factor difference and sum of cubes.


Factor polynomials that require multiple
factoring processes or steps.
Factor trinomials that are quadratic in form.
3
Perfect Square Trinomials
4
Perfect Square Trinomials
A perfect square trinomial comes as a result of squaring a
binomial. Using the distributive property and combining like
terms, we get the following:
(a + b)2 = (a + b)(a + b)
= a2 + ab + ab + b2
= a2 + 2ab + b2
(a – b)2 = (a – b)(a – b)
= a2 – ab – ab + b2
= a2 – 2ab + b2
5
Perfect Square Trinomials
If we look at the resulting trinomials, we can see a pattern
for a perfect square trinomial.
The first and last terms of the trinomial will be perfect
squares, and the middle term will be either plus or minus
twice the product of the first and last terms that are being
squared.
For factoring purposes, the relationship is better seen as
follows.
a2 + 2ab + b2 = (a + b)2
a2 – 2ab + b2 = (a – b)2
6
Example 1 – Factor perfect square trinomials
Factor the following:
a. x2 + 6x + 9
b. m2 – 14m + 49
c. 4a2 + 20a + 25
d. 9x2 – 24xy + 16y2
7
Example 1(a) – Solution
cont’d
First, we must confirm that x2 + 6x + 9 is a perfect square
trinomial. The first term is x squared, the last term is
3 squared, and the middle term is twice the product of
these two terms (2  3  x).
This trinomial can be factored into a binomial squared.
x2 + 6x + 9 = (x + 3)2
Check the factorization.
(x + 3)2 = (x + 3)(x + 3)
= x2 + 3x + 3x + 9
= x2 + 6x + 9
8
Example 1(b) – Solution
cont’d
The first term of m2 – 14m + 49 is m squared, the last term
is 7 squared, and the middle term is minus twice the
product of these two terms (2  7  m).
m2 – 14m + 49 = (m – 7)2
Check the factorization
(m – 7)2 = (m – 7)(m – 7)
= m2 – 7m – 7m + 49
= m2 – 14m + 49
9
Example 1(c) – Solution
cont’d
The first term of 4a2 + 20a + 25 is 2a squared, the last term
is 5 squared, and the middle term is twice the product of
these two terms (2  2a  5).
4a2 + 20a + 25 = (2a + 5)2
Check the factorization.
(2a + 5)2 = (2a + 5)(2a + 5)
= 4a2 + 10a + 10a + 25
= 4a2 + 20a + 25
10
Example 1(d) – Solution
cont’d
The first term of the 9x2 – 24xy + 16y2 is 3x squared, the
last term is 4y squared, and the middle term is minus twice
the product of these two terms (2  3x  4y).
9x2 – 24xy + 16y2 = (3x – 4y)2
Check the factorization.
(3x – 4y)2 = (3x – 4y)(3x – 4y)
= 9x2 – 12xy – 12xy + 16y2
= 9x2 – 24xy + 16y2
11
Difference of Squares
12
Difference of Squares
A similar pattern that we can use is the difference of two
squares.
When multiplying two binomials, one that is the sum of two
terms and the other that is the difference of those same two
terms, we get the difference of two squares.
(a + b)(a – b) = a2 – ab + ab – b2
= a2 – b2
From this we can see that a binomial made up of the
difference of two perfect squares can be factored by
following the pattern
a2 – b2 = (a + b)(a – b)
13
Difference of Squares
The factorization is the sum of the two terms multiplied by
the difference of the two terms.
Although the difference of two squares is factorable, the
sum of two squares is not factorable. Many students want
to factor the sum of two square as
(a + b)(a + b)
This is not a correct factorization. We can check this by
multiplying the factored form out.
(a + b)(a + b) = a2 + ab + ab + b2
= a2 + 2ab + b2
14
Difference of Squares
We can see that the middle term of the final polynomial
does not add to zero, so the product does not result in the
sum of two squares.
15
Example 2 – Factoring the difference of two squares
Factor the following:
a. x2 – 9
c. x2 + 49
b. 4a2 – 25
d. 9m2 – 16n2
Solution:
a. The expression x2 – 9 is the difference of x squared and
3 squared, so we can factor it using the difference of two
squares pattern.
x2 – 9 = x2 – 32
= (x + 3)(x – 3)
16
Example 2 – Solution
cont’d
Check the factorization.
(x + 3)(x – 3) = x2 – 3x + 3x – 9
= x2 – 9
b. The expression 4a2 – 25 is the difference of 2a squared
and 5 squared, so we can factor it using the difference of
two squares pattern.
4a2 – 25 = (2a)2 – 52
= (2a + 5)(2a – 5)
17
Example 2 – Solution
cont’d
Check the factorization.
(2a + 5)(2a – 5) = 4a2 – 10a + 10a – 25
= 4a2 – 25
c. The expression x2 + 49 is the sum of two squares, not
the difference of two squares, and is not factorable using
the rational numbers.
18
Example 2 – Solution
cont’d
d. The expression 9m2 – 16n2 is the difference of 3m
squared and 4n squared, so we can factor it using the
difference of two squares pattern.
9m2 – 16n2 = (3m)2 – (4n)2
= (3m – 4n)(3m + 4n)
Check the factorization.
(3m – 4n)(3m + 4n) = 9m2 + 12mn – 12mn – 16n2
= 9m2 – 16n2
19
Difference and Sum of Cubes
20
Difference and Sum of Cubes
Remember that the sum of two squares is not factorable;
only the difference of two squares is. Two other patterns
that we can use are the sum and difference of two cubes.
The sign changes in these patterns are very important, and
we should be very careful when using these patterns to get
the signs correct.
Sum of two cubes
Difference of two cubes
21
Example 3 – Factor sum and differences of cubes
Factor the following:
a. x3 + 27
b. m3 – 8
c. 8a3 + 125b3
d. 2p3 – 54r3
22
Example 3(a) – Solution
The first term of x3 + 27 is x cubed, and the second term
is 3 cubed, so we can use the pattern for the sum of two
cubes.
x3 + 27 = x3 + 33
= (x + 3)(x2 – 3x + 9)
Check the factorization.
(x + 3) (x2 – 3x + 9) = x2 – 3x2 + 9x + 3x2 – 9x + 27
= x3 + 27
23
Example 3(b) – Solution
cont’d
The first term of m3 – 8 is m cubed, and the second term is
2 cubed, so we can use the pattern for the difference of two
cubes.
m3 – 8 = m3 – 23
= (m – 2)(m2 + 2m + 4)
Check the factorization.
(m – 2)(m2 + 2m + 4) = m3 + 2m2 + 4m – 2m2 – 4m – 8
= m3 – 8
24
Example 3(c) – Solution
cont’d
The first term of 8a3 + 125b3 is 2a cubed, and the second
term is 5b cubed, so we can use the pattern for the sum of
two cubes.
8a3 + 125b3 = (2a)3 + (5b)3
= (2a + 5b)(4a2 – 10ab + 25b2)
Check the factorization.
(2a + 5b)(4a2 – 10ab + 25b2)
= 8a3 – 20a2b + 50ab2 + 20a2b – 50ab2 + 125b3
= 8a3 + 125b3
25
Example 3(d) – Solution
cont’d
For the expression 2p3 – 54r3, we will first factor out the 2
that is in common and then use the pattern for the
difference of two cubes.
2p3 – 54r3 = 2(p3 – 27r3) = 2[p3 – (3r)3]
= 2(p – 3r)(p2 + 3pr + 9r2)
Check the factorization.
2(p – 3r)(p2 + 3pr + 9r2)
= 2(p3 + 3p2r + 9pr2 – 3p2r – 9pr2 – 27r3)
= 2(p3 – 27r3)
= 2p3 – 54r3
26
Multistep Factorizations
27
Multistep Factorizations
Some polynomials take several steps to factor completely
or require slightly different thinking to find the key to
factoring them.
Always look for the greatest common factor and factor that
out first. Then look for a pattern that you recognize and
begin factoring.
Several of the patterns that we have discussed can be
found in more complicated expressions.
28
Multistep Factorizations
The sum or difference of two cubes or the difference of two
squares can be found in expressions with exponents that
are multiples of 2 or 3.
x6 – y6 = (x3)2 – (y3)2
x6 – y6 = (x2)3 – (y2)3
This expression can be looked at as either a difference of
two squares or a difference of two cubes.
This leads to two different paths to factoring, but they have
the same result in the end. Using the difference of two
squares first may make it easier to factor completely.
29
Multistep Factorizations
Difference of squares, then sum and difference of cubes.
x6 – y6 = (x3)2 – (y3)2
= (x3 + y3)(x3 – y3)
= (x + y)(x2 – xy + y2)(x – y)(x2 + xy + y2)
Difference of cubes, then difference of squares and AC
method.
x6 – y6 = (x2)3 – (y2)3
= (x2 – y2)(x4 + x2y2 + y4)
= (x + y)(x – y)(x2 – xy + y2)(x2 + xy + y2)
30
Example 4 – Multistep factoring
Factor the following:
a. x4 – 81
b. 3x6 – 192y6
Solution:
a. This expression x4 – 81 is the difference of two squares,
since the first term is x2 squared and the second term is
9 squared.
x4 – 81 = (x2)2 – 92
The second factor is still
= (x2 + 9)(x2 – 9)
= (x2 + 9)(x + 3)(x – 3)
the difference of two
squares, so we can factor
again.
31
Example 4 – Solution
cont’d
Check the factorization.
(x2 + 9)(x + 3)(x – 3) = (x2 + 9)(x2 – 3x + 3x – 9)
= (x2 + 9)(x2 – 9)
= x4 – 9x2 + 9x2 – 81
= x4 – 81
b. We first factor out the 3 that is in common. The
remaining expression x6 – 64y6 can be viewed as
either the difference of two squares or the difference of
two cubes.
32
Example 4 – Solution
cont’d
If we consider it to be the difference of two cubes, we
end up needing to factor again, using the difference of
two squares to completely finish the factoring.
3x6 – 192y6 = 3(x6 – 64y6)
= 3[(x2)3 – (4y2)3]
Factor using the
difference of two cubes.
= 3(x2 – 4y2)(x4 + 4x2y2 + 16y4)
= 3(x – 2y)(x + 2y)(x4 + 4x2y2 + 16y4)
Finish by using the difference of two squares.
= 3(x – 2y)(x + 2y)(x2 – 2xy + 4y2)(x2 + 2xy + 4y2)
33
Example 4 – Solution
cont’d
Check the factorization.
3(x – 2y)(x + 2y)(x2 – 2xy + 4y2)(x2 + 2xy + 4y2)
= 3(x2 – 4y2)(x4 + 4x2y2 + 16y4)
= 3(x6 + 4x2y2 + 16x2y2 – 4x4y2 – 16x2y4 – 64y6)
= 3(x6 – 64y6)
= 3x6 – 192y6
34
Trinomials in Quadratic Form
35
Trinomials in Quadratic Form
Trinomials of one variable are said to be in quadratic form
when the degree of the highest term is twice that of the
next term and the final term is a constant.
a(expression)2 + b(expression) + c
These types of situations take practice to see, so consider
the following example and pay close attention to the
thinking behind each factorization.
36
Example 5 – Multistep factoring
Factor the following:
a. 3a6 – a3 – 10
b. 12w4 + 52w2 + 35
Solution:
a. The trinomial 3a6 – a3 – 10 is quadratic in form because
the first term has degree twice that of the second term
and the third term is a constant.
We can use a substitution for a3 to make the trinomial
appear to be quadratic. Then factor and replace the a3
back into the expression.
37
Example 5 – Solution
cont’d
We can use any variable for this substitution, so for this
problem, we will use u. Letting u = a3, we get the
following:
3a6 – a3 – 10 = 3(a3)2 – a3 – 10
= 3u2 – u – 10
Substitute in u and factor.
= (3u + 5)(u – 2)
u = a3
= (3a3 + 5)(a3 – 2)
Replace the u with a3.
Check the factorization.
(3a3 + 5)(a3 – 2) = 3a6 – 6a3 + 5a3 – 10
= 3a6 – a3 – 10
38
Example 5 – Solution
cont’d
b. The trinomial 12w4 + 52w2 + 35 is quadratic in form. Use
substitution and then factor the remaining quadratic.
Using u = w2, we get the following:
12w4 + 52w2 + 35 = 12(w2)2 + 52w2 + 35
= 12u2 + 52u2 + 35
Substitute in u and
factor.
= (2u + 7)(6u + 5)
u = w2
= (2w2 + 7)(6w2 + 5)
Replace the u
with w2.
Check the factorization.
(2w2 + 7)(6w2 + 5) = 12w4 + 10w2 + 42w2 + 35
= 12w4 + 52w2 + 35
39
Choosing a Factoring Method
40
Choosing a Factoring Method
When we are faced with a polynomial to factor, we should
choose a method that is best suited to factor the
polynomial. We have looked at four methods of factoring:
• Factor by grouping
• AC method
• Trial and error
41
Choosing a Factoring Method
• Pattern recognition
• Perfect square trinomials
• Difference of squares
• Difference of cubes
• Sum of cubes
• Quadratic form
42
Choosing a Factoring Method
All of these methods start by requiring you to take out the
greatest common factor of all the terms.
If you do not factor out the GCF, all of these methods
become more complicated or impossible.
Determining what method is best depends on the
characteristics of the polynomial we are trying to factor.
43
Choosing a Factoring Method
Factor by Grouping
Factoring by grouping is used when we have four terms in
the polynomial.
5x2 + 10x – 3x – 6
12a2 – 14a + 30ab – 35b
44
Choosing a Factoring Method
AC method or Trial and Error
Either the AC method or trial and error can be used with
quadratics in standard form ax2 + bx + c.
x2 + 5x + 6
4x2 – 2x – 15
Trial and error is easiest when a = 1. After using the AC
method, many people start to see more of the common
factorization patterns and actually start using more trial and
error.
45
Choosing a Factoring Method
Pattern Recognition
Recognizing patterns can help you to factor some
polynomials quickly. The following types of polynomials can
all be factored by using a pattern.
Perfect square trinomial
Difference of squares
25x2 – 16
x2 + 6x + 9
Sum of cubes
Difference of cubes
Quadratic form
x3 + 8
27x3 – 64
x4 – 5x2 – 6
46
Choosing a Factoring Method
Perfect Square Trinomials
A polynomial in the form of a perfect square trinomial
a2 + 2ab + b2 or a2 – 2ab + b2 can be factored by using the
patterns
a2 + 2ab + b2 = (a + b)2
a2 – 2ab + b2 = (a – b)2
Difference of Squares
The difference of two squares a2 – b2 can be factored by
using the pattern
a2 – b2 = (a + b)(a – b)
47
Choosing a Factoring Method
Sum and Difference of Cubes
The sum or difference of cubes a3 + b3 or a3 – b3 can be
factored by using the patterns
a3 + b3 = (a + b)(a2 – ab + b2)
a3 – b3 = (a – b)(a2 + ab + b2)
Quadratic Form
A polynomial that is in quadratic form, that is,
a(expression)2 + b(expression) + c, can be factored by
using either the AC method or trial and error.
48