Section 5.6 Factoring Strategies
INTRODUCTION
Let’s review what you should know about factoring.
(1) “Factors imply multiplication”
Whenever we refer to factors, we are—either directly or indirectly—referring to a product, a
multiplication.
To consider factors of 15, for example, we think of the numbers that multiply to get 15, namely 3 and 5
(or 1 and 15). We’re not looking for numbers that add to get 15, like 7 and 8.
Likewise, when we factor the trinomial x2 + 5x + 6 as (x + 2)(x + 3) we can always check our work
by multiplying, by distributing.
(2) “Factoring is the reverse of the distributive property.”
We can always look for a common monomial factor in all of the terms, something that has been
distributed through to other terms. For example, in the polynomial 2ab3 + 2ac4, you probably see the
common factor of 2a, so we can “factor out” the monomial 2a.
2ab3 + 2ac4 = 2a(b3 + c4)
We can always check such factoring by multiplying (distributing) the 2a through the quantity.
We don’t need to restrict this type of factoring to just binomials. We can factor out common terms
from trinomials (and larger polynomials) as well.
Example 1:
Procedure:
Factor out the greatest common monomial factor from each polynomial.
a)
7x2 – 21y
b)
4x4 + 6x3 – 9x2
c)
5x3 – 20x2 + 10x
d)
2x4y – 8x3y2 + 6x2y3 – 10xy4
a)
Sometimes we can factor
out only a number:
b)
7x2 – 21y = 7(x2 – 3y)
c)
Sometimes we can factor
out both a number and a
variable:
5x3 – 20x2 + 10x
=
Factoring Strategies
5x(x2 – 4x + 2)
Sometimes we can factor out only
a variable:
4x4 + 6x3 – 9x2 = x2(4x2 + 6x – 9)
d)
Sometimes we need to factor out something
common to all four terms:
2x4y – 8x3y2 + 6x2y3 – 10xy4
= 2xy(x3 – 4x2y + 3xy2 – 5y3)
page 5.6 - 1
Exercise 1
Factor out the greatest common monomial factor from each polynomial.
a)
6a3 + 15b2
b)
6c7 + 10c5 – 25c3
c)
8w4 + 12w3 – 36w2
d)
9a4b4 + 6a3b3 – 12a2b2 – 3ab
(4) Some trinomials can be factored into a product of two binomials.
The last two sections introduced you to factoring trinomials using the Factor Game:
•
In Section 5.4, you learned to rewrite the trinomial so that it had four terms. This new four-term
polynomial could then be factored using Factor By Grouping.
•
In Section 5.5 you learned a one-step method for trinomials that have a lead coefficient of 1.
•
In Section 5.5 you also learned to factor perfect square trinomials.
Example 3:
Factor the trinomial 4x2 + 3x – 10.
Key # = - 40 
 gives a result of + 8 and - 5.
(i)
The Factor Game, based on
(ii)
We can use this result to “split” the
middle term into two terms and use
factor by grouping:
Sum # = + 3 
4x2 + 3x – 10
= 4x2 + 8x – 5x – 10
= (4x2 + 8x) + (- 5x – 10)
= 4x(x + 2) + -5(x + 2)
= (4x – 5)(x + 2)
So, the factors of
Factoring Strategies
4x2 + 3x – 10
are
(4x – 5) and (x + 2).
page 5.6 - 2
Exercise 3
Factor each trinomial.
a)
3x2 + 13x + 12
b)
8w2 + 6w – 5
c)
y2 – 11y + 30
d)
c2 – 2c – 48
FACTORING TWO SPECIAL TYPES OF BINOMIALS
Let us consider two different binomials: x2 – 25 and
common, so we can’t factor out a GCF (except 1).
x2 + 25. In each, the terms have nothing in
Notice that each term, x2 and 25, is a perfect square:
we call x2 – 25 the difference of squares and
x2 + 25 is the sum of squares.
Furthermore, we can write each binomial as a trinomial by adding in 0x as a middle term:
x2 – 25 can be written as x2 + 0x – 25 ; likewise x2 + 25 can be written as x2 + 0x + 25
Factoring Strategies
page 5.6 - 3
Example 4:
Factor each binomial by first writing it as a trinomial with 0x as the middle term.
a)
Procedure:
x2 – 25
x2 + 25
Write each as a trinomial and use the Factor Game.
x2 + 0x – 25
a)
b)
Key # = - 25  gives a result

of - 5 and + 5.
Sum # = 0 
So,
Factors into
Exercise 4
x2 + 0x + 25
b)
x2 + 0x – 25
Key # = + 25 
Sum # = 0
 has no solutions.

but x2 + 0x + 25
is prime!
(x – 5)(x + 5)
Factor each binomial by first writing it as a trinomial with 0x as the middle term.
a)
x2 – 49
b)
x2 + 49
c)
x2 + 81
d)
x2 – 81
Think about it:
What observations and conclusions can be made regarding the binomials and their
factorizations in Exercise 4, above.
Answer: Only the difference of squares binomial is factorable; the sum of squares binomial is prime.
FACTORING THE DIFFERENCE OF SQUARES
Factoring Strategies
page 5.6 - 4
The previous exercise was intended to show you that the difference of squares binomial is factorable, but the
sum of squares binomial is not. Once we understand that, we can factor the difference of squares directly, without
having to rewrite it as a trinomial.
Recall from Section 4.5 that two binomials are conjugates if their first terms are exactly the same but their
second terms are opposites, as in (a + b) and (a – b).
Recall, also, that ...
The product of two conjugates is always the difference of squares and fits this pattern:
(a + b)(a – b) = (a)2 – (b)2
For example,
a)
(x + 7)(x – 7)
b)
= x2 – 49
(3x – 2)(3x + 2)
(5x3 – 4)(5x3 + 4)
c)
= 9x2 – 4
= 25x6 – 16
Each of these products is the difference of squares. Looking at each of these from a factoring point of view,
we can see that ...
x2 – 49
a)
b)
= (x + 7)(x – 7)
9x2 – 4
= (3x – 2)(3x + 2)
25x6 – 16
c)
= (5x3 – 4)(5x3 + 4)
Each of these factors is a pair of conjugates. In other words,
The factorization of the difference of squares is the product of two conjugates and fits this pattern:
(a)2 – (b)2 = (a – b)(a + b)
Caution:
Example 5:
The sum of squares, x2 + b2, is not factorable. In
other words, the sum of squares is prime.
Factor. If the binomial isn’t factorable, write prime.
a)
Procedure:
9x2 – 64
b)
x10 – 49
x2 + 4
c)
Identify, if possible, each as the difference of squares. You may even write out each term as a
quantity squared (as shown below).
a)
Factoring Strategies
9x2 – 64
b)
x10 – 49
= (3x)2 – (8)2
= (x5)2 – (7)2
= (3x – 8)(3x + 8)
= (x5 – 7)(x5 + 7)
c)
x2 + 4
This is the sum of squares
and is prime.
page 5.6 - 5
Exercise 5
Factor. If the binomial isn’t factorable, write prime. You may do these in one step.
a)
x2 – 25
b)
36w2 – 1
c)
16y2 + 49
d)
64a2 – 81b2
e)
x4 + 100
f)
4x6 – y2
THE LAST STRATEGY: FACTORING COMPLETELY
Sometimes a number has more than two prime factors, such as 30 = 2 · 3 · 5. Likewise, some polynomials
have more than two factors, such as x3 – 4x, as shown in Example 6:
x3 – 4x completely.
Example 6:
Factor
Procedure:
Factor out the GCF and then see if the binomial factor can, itself, be factored.
Answer:
First, we can factor x from each term ................
x3 – 4x
= x(x2 – 4)
(x2
Next, the binomial factor
– 4) can be factored
further because it is the difference of squares ..........
Notice that this polynomial has three factors:
all three are written in the final factorization.
= x(x – 2)(x + 2)
x, (x – 2) and (x + 2), and
When we factor 30 as 2 · 15, this is not a complete prime factorization.
Likewise, when we factor x3 – 4x as x(x2 – 4), it is not completely factored. Example 7 shows us what it
means to factor completely. The idea of factoring completely is the final guideline in the factoring strategy:
Factoring Strategies
page 5.6 - 6
THE FACTORING STRATEGY
0.
If necessary, write the polynomial in descending order.
1.
If possible, factor out a common monomial factor (as large as possible).
If the lead term is negative, factor out - 1 with the GCF.
2.
If the polynomial is a:
a)
binomial, look for the difference of squares (the sum of squares is prime)
b)
trinomial, use the Factor Game to see if it is factorable.
c)
four-term polynomial, try Factor by Grouping.
3.
Factor completely: see if any of the factors found can be factored further.
(Make sure that all factors are represented in the final answer.)
4.
If the original polynomial is not factorable in any of these ways, then it is prime.
Example 7:
Factor each polynomial using the guidelines set forth above.
a)
Procedure:
a)
b)
3x4 + 15x3 – 18x2
b)
4x2 – 6x + 10
First, check for a common monomial that can be factored out.
Second, see if any of the factors can, themselves, be factored further.
Third, make sure that all factors are shown in the final factored form.
Factor out 3x2 from each term:
3x4 + 15x3 – 18x2
Use the Factor Game on the trinomial x2 + 5x – 6.
= 3x2(x2 + 5x – 6)
You can use another piece of paper to show that
x2 + 5x – 6 = (x + 6)(x – 1). So, we get:
= 3x2(x + 6)(x – 1)
Factor out 2 from each term:
Using the Factor Game we’ll find

that 2x2 – 3x + 5 is prime
so we can’t factor any further: 
Factoring Strategies
4x2 – 6x + 10
= 2(2x2 – 3x + 5)
page 5.6 - 7
Example 8:
a)
b)
c)
Factor each polynomial using the guidelines set forth above.
a)
5x3 – 30x2 + 45x
b)
3x + 28 – x2
c)
5x3 – 45x
d)
2x4 + 8x2
Factor out 5x from each term:
5x3 – 30x2 + 45x
Use the Factor Game on the trinomial x2 – 6x + 9.
= 5x(x2 – 6x + 9)
x2 – 6x + 9 is a perfect square: (x – 3)2. So, we get:
= 5x(x – 3)2
First, write the trinomial in descending order:
- x2 + 3x + 28
The lead term is negative, and there is no other GCF,
so factor out - 1
= - 1(x2 – 3x – 28)
x2 – 3x – 28 = (x – 7)(x + 4). So, we get:
= - 1(x – 7)(x + 4)
Factor out 5x from each term:
5x3 – 45x
= 5x(x2 – 9)
x2 – 9 is the difference of squares. So, we get:
d)
Factor out 2x2 from each term:
x2 + 4 is the sum of squares 

which is prime. So, we

can’t factor any further:
Factoring Strategies
= 5x(x – 3)(x + 3)
2x4 + 8x2
= 2x2(x2 + 4)
page 5.6 - 8
Exercise 6
Factor each of these using the guidelines. Be sure to see if a polynomial can be factored more
than once. If the polynomial cannot be factored at all, write prime.
a)
x3 – 13x2 – 30x
b)
9x2 – 30x + 24
c)
5x3 + 20x
d)
10x2 – 4x – 3
e)
y4 – 9y2
f)
8c3 – 50c
g)
- 2x3 + 22x2 – 36x
h)
x2 – 20x + x3
Factoring Strategies
page 5.6 - 9
Answers to each Exercise
Section 5.6
a)
3(2a3 + 5b2)
b)
c3(6c4 + 10c2 – 25)
c)
4w2(2w2 + 3w – 9)
d)
3ab(3a3b3 + 2a2b2 – 4ab – 1)
Exercise 2:
a)
(5a2 + 2)(a – 3)
b)
(4a – b2)(3a + 2b)
Exercise 3:
a)
(3x + 4)(x + 3)
b)
(2w – 1)(4w + 5)
c)
(y – 5)(y – 6)
d)
(c – 8)(c + 6)
a)
(x – 7)(x + 7)
d)
(x – 9)(x + 9)
a)
(x – 5)(x + 5)
d)
(8a – 9b)(8a + 9b)
a)
Exercise 1:
Exercise 4:
Exercise 5:
Exercise 6:
Factoring Strategies
b)
prime
b)
c)
(6w – 1)(6w + 1)
c)
prime
(2x3 – y)(2x3 + y)
prime
f)
x(x – 15)(x + 2)
b)
3(3x – 4)(x – 2)
c)
5x(x2 + 4)
d)
prime
e)
y2(y
f)
2c(2c + 5)(2c – 5)
g)
- 2x(x – 9)(x – 2)
h)
x(x + 5)(x – 4)
+ 3)(y – 3)
e)
prime
page 5.6 - 10
Section 5.6
1.
Focus Exercises
Factor out the greatest common monomial factor from each polynomial.
a)
8x2 – 12xy2
b)
5p5 – 15p4 + 20p2
c)
- 14m3 + 7m
d)
2x3y – 4x2y2 – 8xy3 + 2xy
2.
Factor each polynomial using Factor By Grouping.
a)
6x3 – 15x2 + 10x – 25
b)
8a2 – 12ab + 2a – 3b
c)
4y3 – 4y2 – 3y + 3
d)
2a3 – 5a2b + 2ab – 5b2
3.
Factor each trinomial.
a)
3x2 + 8x + 4
b)
2x2 – 7x + 6
c)
3x2 + 14x – 5
d)
3x2 – 7x – 6
Factoring Strategies
page 5.6 - 11
4.
Factor each trinomial.
a)
x2 + 6x + 5
b)
x2 – x – 6
c)
x2 + 3x – 4
d)
x2 – 2x – 3
e)
x2 + 5x – 6
f)
x2 – 2x – 1
g)
x2 + 2x – 8
h)
x2 – 3x + 2
i)
x2 + 11x + 10
j)
x2 – 6x + 9
5.
Factor. If the binomial isn’t factorable, write prime. You may do these in one step.
a)
x2 – 36
b)
16w2 – 1
c)
25y2 + 81
d)
9a2 + 4b2
e)
x4 – 25
f)
9x6 – 4y2
Factoring Strategies
page 5.6 - 12
6.
Factor. If the polynomial isn’t factorable, write prime.
a)
x2 + 8x – 20
b)
6p2 + 9p
c)
x2 + 12x + 24
d)
4a2 + 49a
e)
x2 + 9y2
f)
x3 + 25x
g)
x2 – 4x + 4
h)
m2 + 8m – 16
7.
Factor each of these using the guidelines. Each can be factored more than once.
a)
x3 – 7x2 – 8x
b)
5x2 – 15x – 20
c)
12x3 – 27x
d)
y4 – 81
e)
- 6x3 – 12x2 + 18x
f)
- 6x2 + 4x + 10
Factoring Strategies
page 5.6 - 13