Factor Special Products · Intermediate Algebra

We have seen that some binomials and trinomials result from special products—squaring binomials and multiplying conjugates. If you learn to recognize these kinds of polynomials, you can use the special products patterns to factor them much more quickly.

Factor Perfect Square Trinomials

Some trinomials are perfect squares. They result from multiplying a binomial times itself. We squared a binomial using the Binomial Squares pattern in a previous chapter.

is called a perfect square trinomial. It is the square of the binomial

3 x + 4 .

In this chapter, you will start with a perfect square trinomial and factor it into its prime factors.

You could factor this trinomial using the methods described in the last section, since it is of the form

a x^{2} + b x + c .

But if you recognize that the first and last terms are squares and the trinomial fits the perfect square trinomials pattern, you will save yourself a lot of work.

To make use of this pattern, you have to recognize that a given trinomial fits it. Check first to see if the leading coefficient is a perfect square,

a^{2} .

The sign of the middle term determines which pattern we will use. When the middle term is negative, we use the pattern

a^{2} - 2 a b + b^{2},

Remember the first step in factoring is to look for a greatest common factor. Perfect square trinomials may have a GCF in all three terms and it should be factored out first. And, sometimes, once the GCF has been factored, you will recognize a perfect square trinomial.

Factor: $100 x^{2} y - 80 x y + 16 y .$


{: valign=”top”}	Is there a GCF? Yes, $4 y,$

so factor it out. | | {: valign=”top”}| Is this a perfect square trinomial? | | {: valign=”top”}| Verify the pattern. | | {: valign=”top”}| Factor. | | {: valign=”top”}{: .unnumbered .unstyled summary=”Is there a GCF in 100 x squared y minus 80xy plus 16y? Yes. Factoring it out, we get 4y open parentheses 25 x squared minus 20x plus 4 close parentheses. Is this a perfect square trinomial? To verify the pattern, we rewrite as 4y open bracket open parentheses 5x close parentheses squared minus 2 times 5x times 2 plus 2 squared close bracket. Factor to get 4y open parentheses 5x minus 2 close parentheses squared. Finally, we check by multiplying.” data-label=””}

Remember: Keep the factor 4y in the final product.

Check:* * *

\begin{matrix} 4 y {(5 x - 2)}^{2} \\ 4 y [{(5 x)}^{2} - 2 \cdot 5 x \cdot 2 + 2^{2}] \\ 4 y (25 x^{2} - 20 x + 4) \\ 100 x^{2} y - 80 x y + 16 y ✓ \end{matrix}

Factor Differences of Squares

The other special product you saw in the previous chapter was the Product of Conjugates pattern. You used this to multiply two binomials that were conjugates. Here’s an example:

Remember, “difference” refers to subtraction. So, to use this pattern you must make sure you have a binomial in which two squares are being subtracted.

It is important to remember that sums of squares do not factor into a product of binomials. There are no binomial factors that multiply together to get a sum of squares. After removing any GCF, the expression

a^{2} + b^{2}

As always, you should look for a common factor first whenever you have an expression to factor. Sometimes a common factor may “disguise” the difference of squares and you won’t recognize the perfect squares until you factor the GCF.

Also, to completely factor the binomial in the next example, we’ll factor a difference of squares twice!

The next example has a polynomial with 4 terms. So far, when this occurred we grouped the terms in twos and factored from there. Here we will notice that the first three terms form a perfect square trinomial.

Factor: $x^{2} - 6 x + 9 - y^{2} .$

Notice that the first three terms form a perfect square trinomial.

| | | {: valign=”top”}| Factor by grouping the first three terms. | | {: valign=”top”}| Use the perfect square trinomial pattern. | | {: valign=”top”}| Is this a difference of squares? Yes. | | {: valign=”top”}| Yes—write them as squares. | | {: valign=”top”}| Factor as the product of conjugates. | | {: valign=”top”}| | | {: valign=”top”}{: .unnumbered .unstyled summary=”We have x squared minus 6x plus 9 minus y squared. We factor by grouping the first three terms. Use the perfect square trinomial pattern to get open parentheses x minus 3 close parentheses squared minus y squared. Is this a difference of squares? Yes. Factor as the product of conjugates, open parentheses x minus 3 minus y close parentheses open parentheses x minus 3 plus y close parentheses. You may want to rewrite the solution as open parentheses x minus y minus 3 close parentheses open parentheses x plus y minus 3 close parentheses.” data-label=””}

You may want to rewrite the solution as $(x - y - 3) (x + y - 3) .$

Factor Sums and Differences of Cubes

There is another special pattern for factoring, one that we did not use when we multiplied polynomials. This is the pattern for the sum and difference of cubes. We will write these formulas first and then check them by multiplication.

| |

| {: valign=”top”}| Distribute. |

| {: valign=”top”}| Multiply. |

| {: valign=”top”}| Combine like terms. |

| {: valign=”top”}{: .unnumbered .unstyled summary=”Open parentheses a plus b close parentheses open parentheses a squared minus ab plus b squared. Distribute: a open parentheses a squared minus ab plus b squared plus b open parentheses a squared minus ab plus b squared close parentheses. Multiply: a cubed minus a squared b plus ab squared plus a squared b minus ab squared plus b cubed. Combine like terms: a cubed plus b cubed.” data-label=””}

The two patterns look very similar, don’t they? But notice the signs in the factors. The sign of the binomial factor matches the sign in the original binomial. And the sign of the middle term of the trinomial factor is the opposite of the sign in the original binomial. If you recognize the pattern of the signs, it may help you memorize the patterns.

a cubed plus b cubed is open parentheses a plus b close parentheses open parentheses a squared minus ab plus b squared close parentheses. a cubed minus b cubed is open parentheses a minus close parentheses open parentheses a squared plus ab plus b squared close parentheses. In both cases, the sign of the first term on the right side of the equation is the same as the sign on the left side of the equation and the sign of the second term is the opposite of the sign on the left side.

The trinomial factor in the sum and difference of cubes pattern cannot be factored.

It be very helpful if you learn to recognize the cubes of the integers from 1 to 10, just like you have learned to recognize squares. We have listed the cubes of the integers from 1 to 10 in [link].

| 1 | 8 | 27 | 64 | 125 | 216 | 343 | 512 | 729 | 1000 | {: valign=”top”}{: summary=”This table has 11 columns and 2 columns. The first column labels each row n and n cubed. The remaining columns of the first row have the numbers 1 through 10. The remaining columns of the second row have the numbers 1, 8, 27, 64, 125, 216, 343, 512, 729, 1000.”}

How to Factor the Sum or Difference of Cubes

Factor: $x^{3} + 64 .$

![Step 1 is to check if the binomial fits the sum or difference of cubes pattern. For this, we check whether it is a sum or difference. x cubed plus 64 is a sum. Next we check if the first and last terms are perfect cubes. They are](/algebra-intermediate-book/resources/CNX_IntAlg_Figure_06_03_012a.jpg) ![Step 2 is to rewrite as cubes. So we rewrite as x cubed plus 4 cubed.](/algebra-intermediate-book/resources/CNX_IntAlg_Figure_06_03_012b.jpg) ![Step 3 is to use either the sum or difference of cubes pattern. Since this is a sum of cubes, we get open parentheses x plus 4 close parentheses open parentheses x squared minus 4x plus 4 squared.](/algebra-intermediate-book/resources/CNX_IntAlg_Figure_06_03_012c.jpg) ![Step 4 is to simplify inside the parentheses. It is already simplified](/algebra-intermediate-book/resources/CNX_IntAlg_Figure_06_03_012d.jpg) ![Step 5 is to check by multiplying the factors.](/algebra-intermediate-book/resources/CNX_IntAlg_Figure_06_03_012e.jpg)

In the next example, we first factor out the GCF. Then we can recognize the sum of cubes.

Key Concepts

Practice Makes Perfect

Factor Perfect Square Trinomials

In the following exercises, factor completely using the perfect square trinomials pattern.

16 y^{2} + 24 y + 9

{(4 y + 3)}^{2}

25 v^{2} + 20 v + 4

36 s^{2} + 84 s + 49

{(6 s + 7)}^{2}

49 s^{2} + 154 s + 121

100 x^{2} - 20 x + 1

{(10 x - 1)}^{2}

64 z^{2} - 16 z + 1

25 n^{2} - 120 n + 144

{(5 n - 12)}^{2}

4 p^{2} - 52 p + 169

49 x^{2} + 28 x y + 4 y^{2}

{(7 x + 2 y)}^{2}

25 r^{2} + 60 r s + 36 s^{2}

100 y^{2} - 52 y + 1

(50 y - 1) (2 y - 1)

64 m^{2} - 34 m + 1

10 j k^{2} + 80 j k + 160 j

10 j {(k + 4)}^{2}

64 x^{2} y - 96 x y + 36 y

75 u^{4} - 30 u^{3} v + 3 u^{2} v^{2}

3 u^{2} {(5 u - v)}^{2}

90 p^{4} + 300 p^{4} q + 250 p^{2} q^{2}

Factor Differences of Squares

In the following exercises, factor completely using the difference of squares pattern, if possible.

25 v^{2} - 1

(5 v - 1) (5 v + 1)

169 q^{2} - 1

4 - 49 x^{2}

(7 x - 2) (7 x + 2)

121 - 25 s^{2}

6 p^{2} q^{2} - 54 p^{2}

6 p^{2} (q - 3) (q + 3)

98 r^{3} - 72 r

24 p^{2} + 54

6 (4 p^{2} + 9)

20 b^{2} + 140

121 x^{2} - 144 y^{2}

(11 x - 12 y) (11 x + 12 y)

49 x^{2} - 81 y^{2}

169 c^{2} - 36 d^{2}

(13 c - 6 d) (13 c + 6 d)

36 p^{2} - 49 q^{2}

16 z^{4} - 1

(2 z - 1) (2 z + 1) (4 z^{2} + 1)

m^{4} - n^{4}

162 a^{4} b^{2} - 32 b^{2}

2 b^{2} (3 a - 2) (3 a + 2) (9 a^{2} + 4)

48 m^{4} n^{2} - 243 n^{2}

x^{2} - 16 x + 64 - y^{2}

(x - 8 - y) (x - 8 + y)

p^{2} + 14 p + 49 - q^{2}

a^{2} + 6 a + 9 - 9 b^{2}

(a + 3 - 3 b) (a + 3 + 3 b)

m^{2} - 6 m + 9 - 16 n^{2}

Factor Sums and Differences of Cubes

In the following exercises, factor completely using the sums and differences of cubes pattern, if possible.

x^{3} + 125

(x + 5) (x^{2} - 5 x + 25)

n^{6} + 512

z^{6} - 27

(z^{2} - 3) (z^{4} + 3 z^{2} + 9)

v^{3} - 216

8 - 343 t^{3}

(2 - 7 t) (4 + 14 t + 49 t^{2})

125 - 27 w^{3}

8 y^{3} - 125 z^{3}

(2 y - 5 z) (4 y^{2} + 10 y z + 25 z^{2})

27 x^{3} - 64 y^{3}

216 a^{3} + 125 b^{3}

(6 a + 5 b) (36 a^{2} - 30 a b + 25 b^{2})

27 y^{3} + 8 z^{3}

7 k^{3} + 56

7 (k + 2) (k^{2} - 2 k + 4)

6 x^{3} - 48 y^{3}

2 x^{2} - 16 x^{2} y^{3}

2 x^{2} (1 - 2 y) (1 + 2 y + 4 y^{2})

−2 x^{3} y^{2} - 16 y^{5}

{(x + 3)}^{3} + 8 x^{3}

−9 (x + 1) (x^{2} + 3)

{(x + 4)}^{3} - 27 x^{3}

{(y - 5)}^{3} - 64 y^{3}

− (3 y + 5) (21 y^{2} - 30 y + 25)

{(y - 5)}^{3} + 125 y^{3}

Mixed Practice

In the following exercises, factor completely.

64 a^{2} - 25

(8 a - 5) (8 a + 5)

121 x^{2} - 144

27 q^{2} - 3

3 (3 q - 1) (3 q + 1)

4 p^{2} - 100

16 x^{2} - 72 x + 81

{(4 x - 9)}^{2}

36 y^{2} + 12 y + 1

8 p^{2} + 2

2 (4 p^{2} + 1)

81 x^{2} + 169

125 - 8 y^{3}

(5 - 2 y) (25 + 10 y + 4 y^{2})

27 u^{3} + 1000

45 n^{2} + 60 n + 20

5 {(3 n + 2)}^{2}

48 q^{3} - 24 q^{2} + 3 q

x^{2} - 10 x + 25 - y^{2}

(x + y - 5) (x - y - 5)

x^{2} + 12 x + 36 - y^{2}

{(x + 1)}^{3} + 8 x^{3}

(3 x + 1) (3 x^{2} + 1)

{(y - 3)}^{3} - 64 y^{3}

Writing Exercises

Why was it important to practice using the binomial squares pattern in the chapter on multiplying polynomials?

Answers will vary.

How do you recognize the binomial squares pattern?

Explain why $n^{2} + 25 \neq {(n + 5)}^{2} .$

Use algebra, words, or pictures.

Answers will vary.

Maribel factored $y^{2} - 30 y + 81$

as ${(y - 9)}^{2} .$

Was she right or wrong? How do you know?

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This table has 4 columns 3 rows and a header row. The header row labels each column I can, confidently, with some help and no, I don’t get it. The first column has the following statements: factor perfect square trinomials, factor differences of squares, factor sums and differences of cubes. The remaining columns are blank. ⓑ What does this checklist tell you about your mastery of this section? What steps will you take to improve?


Are the first and last terms perfect squares?
Check the middle term.
Does it match ${(a - b)}^{2} ?$ Yes.
Write as the square of a binomial.
Check by multiplying: $\begin{matrix} {(9 y - 4)}^{2} \\ {(9 y)}^{2} - 2 \cdot 9 y \cdot 4 + 4^{2} \\ 81 y^{2} - 72 y + 16 ✓ \end{matrix}$


This binomial is a difference. The first and last terms are perfect cubes.
Write the terms as cubes.
Use the difference of cubes pattern.
Simplify.
Check by multiplying.	We’ll leave the check to you.


Factor the common factor.
This binomial is a sum The first and last terms are perfect cubes.
Write the terms as cubes.
Use the sum of cubes pattern.
Simplify.


This binomial is a difference. The first and last terms are perfect cubes.
Write the terms as cubes.
Use the difference of cubes pattern.
Simplify.

Check by multiplying.	We’ll leave the check to you.


Test each term to verify the pattern.
Factor.
Check by multiplying. $\begin{matrix} {(6 x + 7 y)}^{2} \\ {(6 x)}^{2} + 2 \cdot 6 x \cdot 7 y + {(7 y)}^{2} \\ 36 x^{2} + 84 x y + 49 y^{2} ✓ \end{matrix}$

n	1	2	3	4	5	6	7	8	9	10
{: valign=”top”}	———-
$n^{3}$