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.

How to Factor Perfect Square Trinomials

Factor: $9 x^{2} + 12 x + 4$ .

Solution

This table gives the steps for factoring 9 x squared +12 x +4. The first step is recognizing the perfect square pattern “a” squared + 2 a b + b squared. This includes, is the first term a perfect square and is the last term a perfect square. The first term can be written as (3 x) squared and the last term can be written as 2 squared. Also, in the first step, the middle term has to be twice “a” times b. This is verified by 2 times 3 x times 2 being 12 x.

The second step is writing the square of the binomial. The polynomial is written as (3 x) squared + 2 times 3 x times 2 + 2 squared. This is factored as (3 x + 2) squared.

The last step is to check with multiplication.

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.

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

Solution

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.
${(9 y - 4)}^{2}$
${(9 y)}^{2} - 2 \cdot 9 y \cdot 4 + 4^{2}$
$81 y^{2} - 72 y + 16 ✓$

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

Factor: $36 x^{2} + 84 x y + 49 y^{2}$ .

Solution


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

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

Solution

Steps to factor 9x^2 + 50x + 25, showing an initial attempt at perfect square trinomials and then successful factorization using the 'ac' method.
	$\begin{matrix} 9 x^{2} + 50 x + 25 \end{matrix}$
Are the first and last terms perfect squares?	$\begin{matrix} {(3 x)}^{2} {(5)}^{2} \end{matrix}$
Check the middle term—is it $2 a b ?$	$\begin{matrix} {(3 x)}^{2}_{↘} \underset{\underset{30 x}{2 (3 x) (5)}}{}_{↙} {(5)}^{2} \end{matrix}$
No! $30 x \neq 50 x$	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.

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

Solution

	$36 x^{2} y - 48 x y + 16 y$
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.	$4 y {(3 x - 2)}^{2}$
Remember: Keep the factor 4y in the final product.
Check.
$4 y {(3 x - 2)}^{2}$
$4 y [{(3 x)}^{2} - 2 \cdot 3 x \cdot 2 + 2^{2}]$
$4 y {(9 x)}^{2} - 12 x + 4$
$36 x^{2} y - 48 x y + 16 y ✓$

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.

How to Factor Differences of Squares

Factor: $x^{2} - 4$ .

Solution

This table gives the steps for factoring x squared minus 4. The first step is identifying the pattern in the binomial including it is a difference. Also, the first and last terms are verified as squares.

The second step is writing the two terms as squares, x squared and 2 squared.

The second step is writing the two terms as squares, x squared and 2 squared. The third step is to write the factoring as a product of the conjugates (x – 2)(x + 2).

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.

Factor: $64 y^{2} - 1$ .

Solution


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.
$(8 y - 1) (8 y + 1)$
$64 y^{2} - 1 ✓$

Factor: $121 x^{2} - 49 y^{2}$ .

Solution

$\begin{matrix} 121 x^{2} - 49 y^{2} \\ Is this a difference of squares? Yes. & {(11 x)}^{2} - {(7 y)}^{2} \\ Factor as the product of conjugates. & (11 x - 7 y) (11 x + 7 y) \\ Check by multiplying. \\ (11 x - 7 y) (11 x + 7 y) \\ 121 x^{2} - 49 y^{2} ✓ \end{matrix}$

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

Factor: $100 - h^{2}$ .

Solution

This table demonstrates the step-by-step process of factoring the difference of squares expression 100 - h^2, including a caution.
	$100 - h^{2}$
Is this a difference of squares? Yes.	${(10)}^{2} - {(h)}^{2}$
Factor as the product of conjugates.	$(10 - h) (10 + h)$
Check by multiplying. $\begin{matrix} (10 - h) (10 + h) \\ 100 - h^{2} ✓ \end{matrix}$
Be careful not to rewrite the original expression as $h^{2} - 100$ .

Factor $h^{2} - 100$ on your own and then notice how the result differs from $(10 - h) (10 + h)$ .

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

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

Solution

This table illustrates the step-by-step factorization of the algebraic expression x^4 - y^4 into (x - y)(x + y)(x^2 + y^2) using the difference of squares method, complete with a verification.
	$x^{4} - y^{4}$
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.

Factor: $8 x^{2} y - 98 y$ .

Solution

Step-by-step factorization of the algebraic expression 8x^2y - 98y, demonstrating GCF and difference of squares methods.
	$8 x^{2} y - 98 y$
Is there a GCF? Yes, 2y—factor it out!	$2 y (4 x^{2} - 49)$
Is the binomial a difference of squares? Yes.	$2 y ({(2 x)}^{2} - {(7)}^{2})$
Factor as a product of conjugates.	$2 y (2 x - 7) (2 x + 7)$
Check by multiplying. $\begin{matrix} 2 y (2 x - 7) (2 x + 7) \\ 2 y [(2 x - 7) (2 x + 7)] \\ 2 y (4 x^{2} - 49) \\ 8 x^{2} y - 98 y ✓ \end{matrix}$

Factor: $6 x^{2} + 96$ .

Solution

Steps demonstrating the factoring of the polynomial 6x^2 + 96, including GCF extraction and identifying unfactorable sums of squares.
	$6 x^{2} + 96$
Is there a GCF? Yes, 6—factor it out!	$6 (x^{2} + 16)$
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 11.

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

How to Factor the Sum or Difference of Cubes

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

Solution

This table gives the steps for factoring x cubed + 64. The first step is to verify the binomial fits the pattern. Also, to check the sign for a sum or difference. This binomial is a sum that fits the pattern.

The second step is to write the terms as cubes, x cubed + 4 cubed.

The third step is follow the pattern for the sum of two cubes, (x + 4)(x squared minus x times 4 + 4 squared).

The fourth step is to simplify, (x + 4)(x squared minus 4 x +16).

The last step is to check the answer with multiplication.

Factor: $x^{3} - 1000$ .

Solution


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.

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

Factor: $512 - 125 p^{3}$ .

Solution


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: $27 u^{3} - 125 v^{3}$ .

Solution


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.

Factor: $5 m^{3} + 40 n^{3}$ .

Solution


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.

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

Solution

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

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

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

Solution

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

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

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

Solution

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

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

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

Solution

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

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

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

Solution

${(7 x - 2 y)}^{2}$

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

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

Solution

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

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

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

Solution

$(8 m - 1)^{2}$

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

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

Solution

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

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

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

Solution

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

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

Factor Differences of Squares

In the following exercises, factor.

$x^{2} - 16$

Solution

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

$n^{2} - 9$

$25 v^{2} - 1$

Solution

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

$169 q^{2} - 1$

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

Solution

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

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

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

Solution

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

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

$4 - 49 x^{2}$

Solution

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

$121 - 25 s^{2}$

$16 z^{4} - 1$

Solution

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

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

$5 q^{2} - 45$

Solution

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

$98 r^{3} - 72 r$

$24 p^{2} + 54$

Solution

$6 (4 p^{2} + 9)$

$20 b^{2} + 140$

Factor Sums and Differences of Cubes

In the following exercises, factor.

$x^{3} + 125$

Solution

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

$n^{3} + 512$

$z^{3} - 27$

Solution

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

$v^{3} - 216$

$8 - 343 t^{3}$

Solution

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

$125 - 27 w^{3}$

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

Solution

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

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

$7 k^{3} + 56$

Solution

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

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

$2 - 16 y^{3}$

Solution

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

$−2 x^{3} - 16 y^{3}$

Mixed Practice

In the following exercises, factor.

$64 a^{2} - 25$

Solution

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

$121 x^{2} - 144$

$27 q^{2} - 3$

Solution

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

$4 p^{2} - 100$

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

Solution

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

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

$8 p^{2} + 2$

Solution

$2 (4 p^{2} + 1)$

$81 x^{2} + 169$

$125 - 8 y^{3}$

Solution

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

$27 u^{3} + 1000$

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

Solution

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

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

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.

Solution

${(2 w + 15)}^{2}$

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 $144 - b^{2}$ . Factor the binomial.

Writing Exercises

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

Solution

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.

Solution

Answers may 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 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?

perfect square trinomials pattern

If a and b are real numbers,

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

difference of squares pattern

If a and b are real numbers,
This image shows the difference of two squares formula, a squared – b squared = (a – b)(a + b). Also, the squares are labeled, a squared and b squared. The difference is shown between the two terms. Finally, the factoring (a – b)(a + b) are labeled as conjugates.

This image shows the difference of two squares formula, a squared – b squared = (a – b)(a + b). Also, the squares are labeled, a squared and b squared. The difference is shown between the two terms. Finally, the factoring (a – b)(a + b) are labeled as conjugates.

sum and difference of cubes pattern

\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}