7.4 Factor Special Products

The strategy for factoring we developed in the last section will guide you as you factor most binomials, trinomials, and polynomials with more than three terms. 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. You can square a binomial by using FOIL, but using the Binomial Squares pattern you saw in a previous chapter saves you a step. Let’s review the Binomial Squares pattern by squaring a binomial using FOIL.

This image shows the FOIL procedure for multiplying (3x + 4) squared. The polynomial is written with two factors (3x + 4)(3x + 4). Then, the terms are 9 x squared + 12 x + 12 x + 16, demonstrating first, outer, inner, last. Finally, the product is written, 9 x squared + 24 x + 16.

The first term is the square of the first term of the binomial and the last term is the square of the last. The middle term is twice the product of the two terms of the binomial.

$\begin{matrix} {(3 x)}^{2} + 2 (3 x \cdot 4) + 4^{2} \\ 9 x^{2} + 24 x + 16 \end{matrix}$

The trinomial 9x² + 24 +16 is called a perfect square trinomial. It is the square of the binomial 3x+4.

We’ll repeat the Binomial Squares Pattern here to use as a reference in factoring.

When you square a binomial, the product is a perfect square trinomial. In this chapter, you are learning to factor—now, 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 ax² + bx + 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.

Here is the pattern—the reverse of the binomial squares pattern.

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, $a^{2}$ . Next check that the last term is a perfect square, $b^{2}$ . Then check the middle term—is it twice the product, 2ab? If everything checks, you can easily write the factors.

The sign of the middle term determines which pattern we will use. When the middle term is negative, we use the pattern $a^{2} - 2 a b + b^{2}$ , which factors to ${(a - b)}^{2}$ .

The steps are summarized here.

We’ll work one now where the middle term is negative.

Example 2

Factor: $81 y^{2} - 72 y + 16$ .

The first and last terms are squares. See if the middle term fits the pattern of a perfect square trinomial. The middle term is negative, so the binomial square would be ${(a - b)}^{2}$ .


Are the first and last terms perfect squares?
Check the middle term.
Does is match ${(a - b)}^{2}$ ? Yes.
Write the square of a binomial.
Check by mulitplying.
$open parenthesis 9 y minus 4 close parenthesis squared$
${(9 y)}^{2} - 2 \cdot 9 y \cdot 4 + 4^{2}$
$81 y^{2} - 72 y + 16 ✓$

The next example will be a perfect square trinomial with two variables.

Example 4

Factor: $9 x^{2} + 50 x + 25$ .

	$\begin{matrix} 9 x^{2} + 50 x + 25 \end{matrix}$
Are the first and last terms perfect squares?	$open parenthesis 3 x close parenthesis squared open parenthesis 5 close parenthesis squared$
Check the middle term—is it $2 a b ?$	$open parenthesis 3 x close parenthesis squared sub ↘ 2 open parenthesis 3 x close parenthesis open parenthesis 5 close parenthesis 30 x sub ↙ open parenthesis 5 close parenthesis squared$
No! $30 x not equal to 50 x$	This does not fit the pattern!
Factor using the “ac” method.	$9 x^{2} + 50 x + 25$
$\begin{matrix} Notice: \begin{matrix} a c \\ 9 \cdot 25 \\ 225 \end{matrix} and \begin{matrix} 5 \cdot 45 = 225 \\ 5 + 45 = 50 \end{matrix} \end{matrix}$
Split the middle term. Factor by grouping.	$\begin{matrix} 9 x^{2} + 5 x + 45 x + 25 \\ x (9 x + 5) + 5 (9 x + 5) \\ (9 x + 5) (x + 5) \end{matrix}$
Check. $\begin{matrix} (9 x + 5) (x + 5) \\ 9 x^{2} + 45 x + 5 x + 25 \\ 9 x^{2} + 50 x + 25 ✓ \end{matrix}$

Remember the very first step in our Strategy for Factoring Polynomials? It was to ask “is there a greatest common factor?” and, if there was, you factor the GCF before going any further. 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.

Example 5

Factor: $36 x^{2} y - 48 x y + 16 y$ .

	$36 x^{2} y - 48 x y + 16 y$
Is there a GCF? Yes, 4y, so factor it out.	$4 y (9 x^{2} - 12 x + 4)$
Is this a perfect square trinomial?
Verify the pattern.
Factor.	$4 y open parenthesis 3 x minus 2 close parenthesis squared$
Remember: Keep the factor 4y in the final product.
Check.
$4 y open parenthesis 3 x minus 2 close parenthesis squared$
$4 y [{(3 x)}^{2} - 2 \cdot 3 x \cdot 2 + 2^{2}]$
$4 y {(9 x)}^{2} - 12 x + 4$
$36 x^{2} y - 48 x y + 16 y ✓$

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:

$\begin{matrix} (3 x - 4) (3 x + 4) \\ 9 x^{2} - 16 \end{matrix}$

Remember, when you multiply conjugate binomials, the middle terms of the product add to 0. All you have left is a binomial, the difference of squares.

Multiplying conjugates is the only way to get a binomial from the product of two binomials.

To factor, we will use the product pattern “in reverse” to factor the difference of squares. A difference of squares factors to a product of conjugates.

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 $a^{2} + b^{2}$ is prime!

Don’t forget that 1 is a perfect square. We’ll need to use that fact in the next example.

Example 7

Factor: $64 y squared minus 1$ .


Is this a difference? Yes.
Are the first and last terms perfect squares?
Yes - write them as squares.
Factor as the product of conjugates.
Check by multiplying.
$(8 y - 1) (8 y + 1)$
$64 y squared minus 1 ✓$

The binomial in the next example may look “backwards,” but it’s still the difference of squares.

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

Example 10

Factor: $x^{4} - y^{4}$ .

	$x^{4} - y^{4}$
Is this a difference of squares? Yes.	${(x^{2})}^{2} - {(y^{2})}^{2}$
Factor it as the product of conjugates.	$(x^{2} - y^{2}) (x^{2} + y^{2})$
Notice the first binomial is also a difference of squares!	$({(x)}^{2} - {(y)}^{2}) (x^{2} + y^{2})$
Factor it as the product of conjugates. The last factor, the sum of squares, cannot be factored.	$(x - y) (x + y) (x^{2} + y^{2})$
Check by multiplying. $\begin{matrix} (x - y) (x + y) (x^{2} + y^{2}) \\ [(x - y) (x + y)] (x^{2} + y^{2}) \\ (x^{2} - y^{2}) (x^{2} + y^{2}) \\ x^{4} - y^{4} ✓ \end{matrix}$

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.

Example 11

Factor: $8 x squared y minus 98 y$ .

	$8 x squared y minus 98 y$
Is there a GCF? Yes, 2y—factor it out!	$2 y open parenthesis 4 x squared minus 49 close parenthesis$
Is the binomial a difference of squares? Yes.	$2 y open parenthesis open parenthesis 2 x close parenthesis squared minus open parenthesis 7 close parenthesis squared close parenthesis$
Factor as a product of conjugates.	$2 y (2 x - 7) (2 x + 7)$
Check by multiplying. $\begin{matrix} 2 y (2 x - 7) (2 x + 7) \\ 2 y [(2 x - 7) (2 x + 7)] \\ 2 y (4 x^{2} - 49) \\ 8 x^{2} y - 98 y ✓ \end{matrix}$

Example 12

Factor: $6 x squared plus 96$ .

	$6 x squared plus 96$
Is there a GCF? Yes, 6—factor it out!	$6 open parenthesis x squared plus 16 close parenthesis$
Is the binomial a difference of squares? No, it is a sum of squares. Sums of squares do not factor!
Check by multiplying. $\begin{matrix} 6 (x^{2} + 16) \\ 6 x^{2} + 96 ✓ \end{matrix}$

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.

$\begin{matrix} a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2}) \\ a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2}) \end{matrix}$

We’ll check the first pattern and leave the second to you.


Distribute.
Multiply.	$a^{3} - a^{2} b + {a b}^{2} + a^{2} b - {a b}^{2} + b^{3}$
Combine like terms.	$a^{3} + b^{3}$

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.

This figure demonstrates the sign patterns in the sum and difference of two cubes. For the sum of two cubes, this figure shows the first two signs are plus and the first and the third signs are opposite, plus minus. The difference of two cubes has the first two signs the same, minus. The first and the third sign are minus plus.

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

It can 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 Table 7.2.

**Table 7.2**
n	1	2	3	4	5	6	7	8	9	10
$n^{3}$	1	8	27	64	125	216	343	512	729	1000

Be careful to use the correct signs in the factors of the sum and difference of cubes.

Example 15

Factor: $512 minus 125 p cubed$ .


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.

Example 16

Factor: $27 u cubed minus 125 v cubed$ .


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.

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

Example 17

Factor: $5 m cubed plus 40 n cubed$ .


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.

Check. To check, you may find it easier to multiply the sum of cubes factors first, then multiply that product by 5. We’ll leave the multiplication for you.

$5 (\overset{}{m + 2 n}) (\overset{}{m^{2} - 2 m n + 4 n^{2}})$

Key Concepts

Factor perfect square trinomials See Example 1.
$\begin{matrix} Step 1. Does the trinomial fit the pattern? & a^{2} + 2 a b + b^{2} & a^{2} - 2 a b + b^{2} \\ Is the first term a perfect square? & {(a)}^{2} & {(a)}^{2} \\ Write it as a square. \\ Is the last term a perfect square? & {(a)}^{2} {(b)}^{2} & {(a)}^{2} {(b)}^{2} \\ Write it as a square. \\ Check the middle term. Is it 2 a b ? & {(a)}^{2}_{↘} \underset{2 \cdot a \cdot b}{}_{↙} {(b)}^{2} & {(a)}^{2}_{↘} \underset{2 \cdot a \cdot b}{}_{↙} {(b)}^{2} \\ Step 2. Write the square of the binomial. & {(a + b)}^{2} & {(a - b)}^{2} \\ Step 3. Check by multiplying. \end{matrix}$
Factor differences of squares See Example 6.
$\begin{matrix} Step 1. Does the binomial fit the pattern? & a^{2} - b^{2} \\ Is this a difference? & ____-____ \\ Are the first and last terms perfect squares? \\ Step 2. Write them as squares. & {(a)}^{2} - {(b)}^{2} \\ Step 3. Write the product of conjugates. & (a - b) (a + b) \\ Step 4. Check by multiplying. \end{matrix}$
Factor sum and difference of cubes To factor the sum or difference of cubes: See Example 13.
1. Does the binomial fit the sum or difference of cubes pattern? Is it a sum or difference? Are the first and last terms perfect cubes?
2. Write them as cubes.
3. Use either the sum or difference of cubes pattern.
4. Simplify inside the parentheses
5. Check by multiplying the factors.

Practice Makes Perfect

Factor Perfect Square Trinomials

In the following exercises, factor.

$16 y^{2} + 24 y + 9$

$open parenthesis 4 y plus 3 close parenthesis squared$

$25 v^{2} + 20 v + 4$

$36 s^{2} + 84 s + 49$

$open parenthesis 6 s plus 7 close parenthesis squared$

$49 s^{2} + 154 s + 121$

$100 x^{2} - 20 x + 1$

$open parenthesis 10 x minus 1 close parenthesis squared$

$64 z^{2} - 16 z + 1$

$25 n^{2} - 120 n + 144$

$open parenthesis 5 n minus 12 close parenthesis squared$

$4 p^{2} - 52 p + 169$

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

$open parenthesis 7 x minus 2 y close parenthesis squared$

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

$25 n^{2} + 25 n + 4$

$(5 n + 4) (5 n + 1)$

$100 y^{2} - 20 y + 1$

$64 m^{2} - 16 m + 1$

$open parenthesis 8 m minus 1 close parenthesis squared$

$100 x^{2} - 25 x + 1$

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

$10 open parenthesis k plus 4 close parenthesis squared$

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

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

$3 u open parenthesis 5 u minus v close parenthesis squared$

$90 p^{3} + 300 p^{2} q + 250 p q^{2}$

Factor Differences of Squares

In the following exercises, factor.

$x squared minus 16$

$(x - 4) (x + 4)$

$n squared minus 9$

$25 v squared minus 1$

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

$169 q squared minus 1$

$121 x squared minus 144 y squared$

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

$49 x squared minus 81 y squared$

$169 c squared minus 36 d squared$

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

$36 p squared minus 49 q squared$

$4 minus 49 x squared$

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

$121 minus 25 s squared$

$16 z^{4} - 1$

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

$m^{4} - n^{4}$

$5 q squared minus 45$

$5 (q - 3) (q + 3)$

$98 r cubed minus 72 r$

$24 p squared plus 54$

$6 open parenthesis 4 p squared plus 9 close parenthesis$

$20 b squared plus 140$

Factor Sums and Differences of Cubes

In the following exercises, factor.

$x cubed plus 125$

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

$n cubed plus 512$

$z cubed minus 27$

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

$v cubed minus 216$

$8 minus 343 t cubed$

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

$125 minus 27 w cubed$

$8 y cubed minus 125 z cubed$

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

$27 x cubed minus 64 y cubed$

$7 k cubed plus 56$

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

$6 x cubed minus 48 y cubed$

$2 minus 16 y cubed$

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

$−2 x cubed minus 16 y cubed$

Mixed Practice

In the following exercises, factor.

$64 a squared minus 25$

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

$121 x squared minus 144$

$27 q squared minus 3$

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

$4 p squared minus 100$

$16 x^{2} - 72 x + 81$

$open parenthesis 4 x minus 9 close parenthesis squared$

$36 y^{2} + 12 y + 1$

$8 p squared plus 2$

$2 open parenthesis 4 p squared plus 1 close parenthesis$

$81 x squared plus 169$

$125 minus 8 y cubed$

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

$27 u cubed plus 1000$

$45 n^{2} + 60 n + 20$

$5 open parenthesis 3 n plus 2 close parenthesis squared$

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

Everyday Math

Landscaping Sue and Alan are planning to put a 15 foot square swimming pool in their backyard. They will surround the pool with a tiled deck, the same width on all sides. If the width of the deck is w, the total area of the pool and deck is given by the trinomial $4 w^{2} + 60 w + 225$ . Factor the trinomial.

$open parenthesis 2 w plus 15 close parenthesis squared$

Home repair The height a twelve foot ladder can reach up the side of a building if the ladder’s base is b feet from the building is the square root of the binomial $144 minus b squared$ . Factor the binomial.

Writing Exercises

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

Answers may vary.

How do you recognize the binomial squares pattern?

Explain why $n^{2} + 25 \neq {(n + 5)}^{2}$ . Use algebra, words, or pictures.

Answers may vary.

Maribel factored $y^{2} - 30 y + 81$ as $open parenthesis y minus 9 close parenthesis squared$ . 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 the following statements all to be preceded by “I can…”. The first row is “factor perfect square trinomials”. The second row is “factor differences of squares”. The third row is “factor sums and differences of cubes”. In the columns beside these statements are the headers, “confidently”, “with some help”, and “no-I don’t get it!”.

ⓑ On a scale of 1–10, how would you rate your mastery of this section in light of your responses on the checklist? How can you improve this?


Test each term to verify the pattern.
Factor.
Check by mulitplying.
$open parenthesis 6 x plus 7 y close parenthesis squared$
${(6 x)}^{2} + 2 \cdot 6 x \cdot 7 y + {(7 y)}^{2}$
$36 x^{2} + 84 x y + 49 y^{2} ✓$