Factor Special Products

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.

In open parentheses 3x plus 4 close parentheses squared, 3x is a and 4 is b. Writing it as a squared plus 2ab plus b squared, we get open parentheses 3x close parentheses squared plus 2 times 3x times 4 plus 4 squared. This is equal to 9 x squared plus 24x plus 16.

The trinomial $9 x^{2} + 24 x + 16$ 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.

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 the product, $2 a b ?$ If everything checks, you can easily write the factors.

How to Factor Perfect Square Trinomials

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

Solution

Step 1 is to check if the trinomial fits the perfect square trinomials pattern, a squared plus 2ab plus b squared. For this we check if the first term is a perfect square. 9 x squared is the square of 3x. Next we check if the last term is a perfect square. 4 is the square of 2. Next we check if the middle term is 2ab. 12 x is twice 3x times 2. Hence we have a perfect square trinomial.

Step 2 is to write this as the square of a binomial. We write it as open parentheses 3x plus 2 close parentheses squared.

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 it match ${(a - b)}^{2} ?$ Yes.
Write as the square of a binomial.
Check by multiplying: $\begin{matrix} {(9 y - 4)}^{2} \\ {(9 y)}^{2} - 2 \cdot 9 y \cdot 4 + 4^{2} \\ 81 y^{2} - 72 y + 16 ✓ \end{matrix}$

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 multiplying. $\begin{matrix} {(6 x + 7 y)}^{2} \\ {(6 x)}^{2} + 2 \cdot 6 x \cdot 7 y + {(7 y)}^{2} \\ 36 x^{2} + 84 x y + 49 y^{2} ✓ \end{matrix}$

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

Solution


Is there a GCF? Yes, $4 y,$ so factor it out.
Is this a perfect square trinomial?
Verify the pattern.
Factor.

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:

We have open parentheses 3x minus 4 close parentheses open parentheses 3x plus 4. This is of the form a minus b, a plus b. We rewrite as open parentheses 3x close parentheses squared minus 4 squared. Here, 3x is a and 4 is b. This is equal to 9 x squared minus 16.

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 a Binomial Using the Difference of Squares

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

Solution

Step 1 is to check if the binomial 64 y squared minus 1 fits the pattern. For that we check the following: Is this a difference? Yes. Are the first and last terms perfect squares? Yes.

Step 2 is to write both terms as squares, So, we have open parentheses 8y close parentheses squared minus 1 squared.

Step 3 is to write the product of conjugates 8y minus 1, 8y plus 1.

Step 4 is to check. We multiply to get the original binomial

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!

The next example shows variables in both terms.

Factor: $144 x^{2} - 49 y^{2} .$

Solution

Step-by-step factorization of a difference of squares polynomial, including identification, factoring, and verification.
	$144 x^{2} - 49 y^{2}$
Is this a difference of squares? Yes.	${(12 x)}^{2} - {(7 y)}^{2}$
Factor as the product of conjugates.	$(12 x - 7 y) (12 x + 7 y)$
Check by multiplying. $\begin{matrix} (12 x - 7 y) (12 x + 7 y) \\ 144 x^{2} - 49 y^{2} ✓ \end{matrix}$

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.

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

Factor: $48 x^{4} y^{2} - 243 y^{2} .$

Solution

Step-by-step factorization of the polynomial 48x^4y^2 - 243y^2, demonstrating GCF and difference of squares methods, with a final verification.
	$48 x^{4} y^{2} - 243 y^{2}$
Is there a GCF? Yes, $3 y^{2}$ —factor it out!	$3 y^{2} (16 x^{4} - 81)$
Is the binomial a difference of squares? Yes.	$3 y^{2} ({(4 x^{2})}^{2} - {(9)}^{2})$
Factor as a product of conjugates.	$3 y^{2} (4 x^{2} - 9) (4 x^{2} + 9)$
Notice the first binomial is also a difference of squares!	$3 y^{2} ({(2 x)}^{2} - {(3)}^{2}) (4 x^{2} + 9)$
Factor it as the product of conjugates.	$3 y^{2} (2 x - 3) (2 x + 3) (4 x^{2} + 9)$
The last factor, the sum of squares, cannot be factored.
Check by multiplying:
$3 y^{2} (2 x - 3) (2 x + 3) (4 x^{2} + 9)$
$3 y^{2} (4 x^{2} - 9) (4 x^{2} + 9)$
$3 y^{2} (16 x^{4} - 81)$
$48 x^{4} y^{2} - 243 y^{2} ✓$

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

Solution

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


Factor by grouping the first three terms.
Use the perfect square trinomial pattern.
Is this a difference of squares? Yes.
Yes—write them as squares.
Factor as the product of conjugates.

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.

\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.
Combine like terms.

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 will 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 8.

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

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

Step 2 is to rewrite as cubes. So we rewrite as x cubed plus 4 cubed.

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.

Step 4 is to simplify inside the parentheses. It is already simplified

Step 5 is to check by multiplying the factors.

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: $6 x^{3} y + 48 y^{4} .$

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 $6 y .$ We’ll leave the multiplication for you.

The first term in the next example is a binomial cubed.

Factor: ${(x + 5)}^{3} - 64 x^{3} .$

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.

Key Concepts

Perfect Square Trinomials Pattern: If a and b are real numbers,

$\begin{matrix} a^{2} + 2 a b + b^{2} = {(a + b)}^{2} \\ a^{2} - 2 a b + b^{2} = {(a - b)}^{2} \end{matrix}$
How to factor perfect square trinomials.
$\begin{matrix} Step 1. & Does the trinomial fit the pattern? & a^{2} + 2 a b + b^{2} & a^{2} - 2 a b + b^{2} \\ Is the first term a perfect square? & {(a)}^{2} & {(a)}^{2} \\ Write it as a square. \\ Is the last term a perfect square? & {(a)}^{2} {(b)}^{2} & {(a)}^{2} {(b)}^{2} \\ Write it as a square. \\ Check the middle term. Is it 2 a b ? & {(a)}^{2}_{↘} \underset{2 \cdot a \cdot b}{}_{↙} {(b)}^{2} & {(a)}^{2}_{↘} \underset{2 \cdot a \cdot b}{}_{↙} {(b)}^{2} \\ Step 2. & Write the square of the binomial. & {(a + b)}^{2} & {(a - b)}^{2} \\ Step 3. & Check by multiplying. \end{matrix}$
Difference of Squares Pattern: If $a, b$ are real numbers,
How to factor differences of squares.
$\begin{matrix} Step 1. & Does the binomial fit the pattern? & a^{2} - b^{2} \\ Is this a difference? & ____-____ \\ Are the first and last terms perfect squares? \\ Step 2. & Write them as squares. & {(a)}^{2} - {(b)}^{2} \\ Step 3. & Write the product of conjugates. & (a - b) (a + b) \\ Step 4. & Check by multiplying. \end{matrix}$
Sum and Difference of Cubes Pattern
$\begin{matrix} a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2}) \\ a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2}) \end{matrix}$
How to factor the sum or difference of cubes.
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 completely using the perfect square trinomials pattern.

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

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

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

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

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

Solution

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

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

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

Solution

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

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

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

Solution

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

$90 p^{4} + 300 p^{3} q + 250 p^{2} q^{2}$

Factor Differences of Squares

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

$25 v^{2} - 1$

Solution

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

$169 q^{2} - 1$

$4 - 49 x^{2}$

Solution

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

$121 - 25 s^{2}$

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

Solution

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

$98 r^{3} - 72 r$

$24 p^{2} + 54$

Solution

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

$20 b^{2} + 140$

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

Solution

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

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

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

Solution

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

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

$16 z^{4} - 1$

Solution

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

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

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

Solution

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

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

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

Solution

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

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

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

Solution

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

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

Factor Sums and Differences of Cubes

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

$x^{3} + 125$

Solution

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

$n^{6} + 512$

$z^{6} - 27$

Solution

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

$v^{3} - 216$

$8 - 343 t^{3}$

Solution

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

$125 - 27 w^{3}$

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

Solution

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

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

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

Solution

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

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

$7 k^{3} + 56$

Solution

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

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

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

Solution

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

$−2 x^{3} y^{2} - 16 y^{5}$

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

Solution

$9 (x + 1) (x^{2} + 3)$

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

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

Solution

$− (3 y + 5) (21 y^{2} - 30 y + 25)$

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

Mixed Practice

In the following exercises, factor completely.

$64 a^{2} - 25$

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

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

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

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

Solution

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

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

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

Solution

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

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

Writing Exercises

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

Solution

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.

Solution

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?