# 1.2.8: Factoring Special Products

$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$

$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$

$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$

( \newcommand{\kernel}{\mathrm{null}\,}\) $$\newcommand{\range}{\mathrm{range}\,}$$

$$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$

$$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$

$$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$

$$\newcommand{\Span}{\mathrm{span}}$$

$$\newcommand{\id}{\mathrm{id}}$$

$$\newcommand{\Span}{\mathrm{span}}$$

$$\newcommand{\kernel}{\mathrm{null}\,}$$

$$\newcommand{\range}{\mathrm{range}\,}$$

$$\newcommand{\RealPart}{\mathrm{Re}}$$

$$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$

$$\newcommand{\Argument}{\mathrm{Arg}}$$

$$\newcommand{\norm}[1]{\| #1 \|}$$

$$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$

$$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$

$$\newcommand{\vectorA}[1]{\vec{#1}} % arrow$$

$$\newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow$$

$$\newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$

$$\newcommand{\vectorC}[1]{\textbf{#1}}$$

$$\newcommand{\vectorD}[1]{\overrightarrow{#1}}$$

$$\newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}}$$

$$\newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}}$$

$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$

$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$

$$\newcommand{\avec}{\mathbf a}$$ $$\newcommand{\bvec}{\mathbf b}$$ $$\newcommand{\cvec}{\mathbf c}$$ $$\newcommand{\dvec}{\mathbf d}$$ $$\newcommand{\dtil}{\widetilde{\mathbf d}}$$ $$\newcommand{\evec}{\mathbf e}$$ $$\newcommand{\fvec}{\mathbf f}$$ $$\newcommand{\nvec}{\mathbf n}$$ $$\newcommand{\pvec}{\mathbf p}$$ $$\newcommand{\qvec}{\mathbf q}$$ $$\newcommand{\svec}{\mathbf s}$$ $$\newcommand{\tvec}{\mathbf t}$$ $$\newcommand{\uvec}{\mathbf u}$$ $$\newcommand{\vvec}{\mathbf v}$$ $$\newcommand{\wvec}{\mathbf w}$$ $$\newcommand{\xvec}{\mathbf x}$$ $$\newcommand{\yvec}{\mathbf y}$$ $$\newcommand{\zvec}{\mathbf z}$$ $$\newcommand{\rvec}{\mathbf r}$$ $$\newcommand{\mvec}{\mathbf m}$$ $$\newcommand{\zerovec}{\mathbf 0}$$ $$\newcommand{\onevec}{\mathbf 1}$$ $$\newcommand{\real}{\mathbb R}$$ $$\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}$$ $$\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}$$ $$\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}$$ $$\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}$$ $$\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$$ $$\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$$ $$\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$$ $$\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$$ $$\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}$$ $$\newcommand{\laspan}[1]{\text{Span}\{#1\}}$$ $$\newcommand{\bcal}{\cal B}$$ $$\newcommand{\ccal}{\cal C}$$ $$\newcommand{\scal}{\cal S}$$ $$\newcommand{\wcal}{\cal W}$$ $$\newcommand{\ecal}{\cal E}$$ $$\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}$$ $$\newcommand{\gray}[1]{\color{gray}{#1}}$$ $$\newcommand{\lgray}[1]{\color{lightgray}{#1}}$$ $$\newcommand{\rank}{\operatorname{rank}}$$ $$\newcommand{\row}{\text{Row}}$$ $$\newcommand{\col}{\text{Col}}$$ $$\renewcommand{\row}{\text{Row}}$$ $$\newcommand{\nul}{\text{Nul}}$$ $$\newcommand{\var}{\text{Var}}$$ $$\newcommand{\corr}{\text{corr}}$$ $$\newcommand{\len}[1]{\left|#1\right|}$$ $$\newcommand{\bbar}{\overline{\bvec}}$$ $$\newcommand{\bhat}{\widehat{\bvec}}$$ $$\newcommand{\bperp}{\bvec^\perp}$$ $$\newcommand{\xhat}{\widehat{\xvec}}$$ $$\newcommand{\vhat}{\widehat{\vvec}}$$ $$\newcommand{\uhat}{\widehat{\uvec}}$$ $$\newcommand{\what}{\widehat{\wvec}}$$ $$\newcommand{\Sighat}{\widehat{\Sigma}}$$ $$\newcommand{\lt}{<}$$ $$\newcommand{\gt}{>}$$ $$\newcommand{\amp}{&}$$ $$\definecolor{fillinmathshade}{gray}{0.9}$$
##### Learning Objectives

By the end of this section, you will be able to:

• Factor perfect square trinomials
• Factor differences of squares
• Factor sums and differences of cubes
##### Be Prepared

Before you get started, take this readiness quiz.

1. Simplify $$(3x^2)^3$$.

2. Multiply $$(m+4)^2$$.

3. Multiply $$(x−3)(x+3)$$.

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

## Factoring Differences of Squares

One special product we are familiar with is the Product of Conjugates pattern. We use this to multiply two binomials that were conjugates. Here’s an example:

$(2x-5)(2x+5)=4x^2-25.\nonumber$

A difference of squares factors to a product of conjugates (in this context, $$a-b$$ is said to be conjugate to $$a+b$$, and vice versa.)

Difference of Squares Pattern

If $$a$$ and $$b$$ are (or represent) real numbers,

$a^2-b^2=(a+b)(a-b).\nonumber$

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

##### Example $$\PageIndex{1}$$

Factor $$64y^2−1$$.

###### Solution

Note that $$(8y)^2=64y^2$$ and $$1^2=1$$, so that

\begin{align*} &64y^2-1\\ = &(8y)^2-(1)^2 \\ =&(8y+1)(8y-1).\end{align*}

Try It $$\PageIndex{2}$$

Factor $$121m^2−1$$.

$$(11m−1)(11m+1)$$

Try It $$\PageIndex{3}$$

Factor $$81y^2−1$$.

$$(9y−1)(9y+1)$$

Factor Difference of Squares

$$\begin{array} {llll} \textbf{Step 1.} &\text{Does the binomial fit the pattern?} &\qquad &\hspace{5mm} a^2−b^2 \\ &\text{Is this a difference?} &\qquad &\hspace{2mm} \text{____−____} \\ &\text{Are the first and last terms perfect squares?} & & \\ \textbf{Step 2.} &\text{Write them as squares.} &\qquad &\hspace{3mm} (a)^2−(b)^2 \\ \textbf{Step 3.} &\text{Write the product of conjugates.} &\qquad &(a−b)(a+b) \\ \textbf{Step 4.} &\text{Check by multiplying.} & & \end{array}$$

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 $$a^2+b^2$$ can not be factored any further!

The next example shows variables in both terms.

##### Example $$\PageIndex{4}$$

Factor $$144x^2−49y^2$$.

###### Solution
 $$\quad 144x^2−49y^2$$ Is this a difference of squares? Yes. $$=(12x)^2−(7y)^2$$ Factor as the product of conjugates. $$=(12x−7y)(12x+7y)$$ Check by multiplying. $$=144x^2−49y^2\checkmark$$ Conclude. The factorization is $$(12x-7y)(12x+7y)$$.

Try It $$\PageIndex{5}$$

Factor $$196m^2−25n^2$$.

$$(16m−5n)(16m+5n)$$

Try It $$\PageIndex{6}$$

Factor $$121p^2−9q^2$$.

$$(11p−3q)(11p+3q)$$

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

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

##### Example $$\PageIndex{7}$$

Factor $$48x^4y^2−243y^2$$.

###### Solution
 $$\quad 48x^4y^2−243y^2$$ Is there a GCF? Yes, $$3y^2$$ - factor it out! $$=3y^2(16x^4−81)$$ Is the binomial a difference of squares? Yes. $$=3y^2\left((4x^2)^2−(9)^2\right)$$ Factor as a product of conjugates. $$=3y^2(4x^2−9)(4x^2+9)$$ Notice the first binomial is also a difference of squares! $$=3y^2((2x)^2−(3)^2)(4x^2+9)$$ Factor it as the product of conjugates. $$=3y^2(2x−3)(2x+3)(4x^2+9)$$ The last factor, the sum of squares, cannot be factored. Check by multiplying: $$\quad 3y^2(2x−3)(2x+3)(4x^2+9)$$ $$=3y^2(4x^2−9)(4x^2+9)$$ $$=3y^2(16x^4−81)$$ $$=48x^4y^2−243y^2\checkmark$$ Conclude. The factorization is $$3y^2(2x-3)(2x+3)(4x^2+9)$$.

Try It $$\PageIndex{8}$$

Factor $$2x^4y^2−32y^2$$.

$$2y^2(x−2)(x+2)(x^2+4)$$

Try It $$\PageIndex{9}$$

Factor $$7a^4c^2−7b^4c^2$$.

$$7c^2(a−b)(a+b)(a^2+b^2)$$

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.

##### Example $$\PageIndex{10}$$

Factor $$x^2−6x+9−y^2$$.

###### Solution

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

 $$x^2−6x+9−y^2$$ Factor by grouping the first three terms. Use the perfect square trinomial pattern. $$(x-3)^{2}-y^{2}$$ Is this a difference of squares? Yes. Yes—write them as squares. (already in that form) Factor as the product of conjugates. $$(x-3-y)(x-3+y)$$ Conclude. The factorization is $$(x−y−3)(x+y−3)$$.

Try It $$\PageIndex{11}$$

Factor $$x^2−10x+25−y^2$$.

$$(x−5−y)(x−5+y)$$

Try It $$\PageIndex{12}$$

Factor $$x^2+6x+9−4y^2$$.

$$(x+3−2y)(x+3+2y)$$

## Factoring Perfect Square Trinomials (optional discussion)

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.

The trinomial $$9x^2+24x+16$$ is called a perfect square trinomial. It is the square of the binomial $$3x+4$$.

In this chapter, we will start with a perfect square trinomial and factor it into its as many factors as possible. We could factor this trinomial using the methods described in the last section, since it is of the form $$ax^2+bx+c$$. But if we recognize that the first and last terms are squares and the trinomial fits the perfect square trinomials pattern, we will save ourselves a lot of work. Here is the pattern—the reverse of the binomial squares pattern.

Perfect Square Trinomial Pattern

If $$a$$ and $$b$$ are (or represent) real numbers

$a^2+2ab+b^2=(a+b)^2,\nonumber$

$a^2−2ab+b^2=(a−b)^2.\nonumber$

To make use of this pattern, we 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, $$2ab$$? If everything checks, we 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 $$a^2−2ab+b^2$$, which factors to $$(a−b)^2$$.

The steps are summarized here.

Factoring Perfect Square Trinomials

$$\begin{array} {lllll} \textbf{Step 1.} &\text{Does the trinomial fit the pattern?} &\quad &\hspace{7mm} a^2+2ab+b^2 &\hspace{7mm} a^2−2ab+b^2 \\ &\text{Are the first and last terms perfect squares?} &\quad & &\\ &\text{Write them as squares.} &\quad &\hspace{5mm}(a)^2\hspace{16mm} (b)^2 &\hspace{6mm}(a)^2\hspace{16mm} (b)^2 \\ &\text{Check the middle term. Is it }2ab? &\quad &\hspace{12mm} {\,}^{\searrow}{\,}_{2·a·b}{\,}^{\swarrow} &\hspace{12mm} {\,}^{\searrow}{\,}_{2·a·b}{\,}^{\swarrow} \\ \textbf{Step 2.} &\text{Write the square of the binomial.} &\quad &\hspace{13mm} (a+b)^2 &\hspace{13mm} (a−b)^2 \\ \textbf{Step 3.} &\text{Check by multiplying.} & & & \end{array}$$

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, we will recognize a perfect square trinomial.

The above is for consideration, but we will not give any examples of this here.

##### Writing Exercises $$\PageIndex{13}$$
1. Why might it be useful to recognize the special products?
2. Explain why $$(x+2)^2\not = x^2+4$$.
3. Give an example of the difference-of-squares pattern with three variables?
4. Use the difference of squares pattern to simplify and factor 49-16.
5. Is there a sum of squares formula?
6. Demonstrate why $$a^2-b^2=(a-b)(a+b)$$.
##### Exit Problem

Factor $$25x^2-16y^6$$.

## Key Concepts

• Difference of Squares Pattern: If $$a$$, $$b$$ are (or represent) real numbers,
$a^2-b^2=(a+b)(a-b)$
• How to factor differences of squares
$$\begin{array} {llll} \textbf{Step 1.} &\text{Does the binomial fit the pattern?} &\qquad &\hspace{5mm} a^2−b^2 \\ &\text{Is this a difference?} &\qquad &\hspace{2mm} \text{____−____} \\ &\text{Are the first and last terms perfect squares?} & & \\ \textbf{Step 2.} &\text{Write them as squares.} &\qquad &\hspace{3mm} (a)^2−(b)^2 \\ \textbf{Step 3.} &\text{Write the product of conjugates.} &\qquad &(a−b)(a+b) \\ \textbf{Step 4.} &\text{Check by multiplying.} & & \end{array}$$
• Perfect Square Trinomials Pattern: If $$a$$ and $$b$$ are (or represent) real numbers,

$\begin{array} {l} a^2+2ab+b^2=(a+b)^2 \\ a^2−2ab+b^2=(a−b)^2\end{array} \nonumber$

• How to factor perfect square trinomials
$$\begin{array} {lllll} \textbf{Step 1.} &\text{Does the trinomial fit the pattern?} &\quad &\hspace{7mm} a^2+2ab+b^2 &\hspace{7mm} a^2−2ab+b^2 \\ &\text{Are the first and last terms perfect squares?} &\quad & &\\ &\text{Write them as squares.} &\quad &\hspace{5mm}(a)^2\hspace{16mm} (b)^2 &\hspace{6mm}(a)^2\hspace{16mm} (b)^2 \\ &\text{Check the middle term. Is it }2ab? &\quad &\hspace{12mm} {\,}^{\searrow}{\,}_{2·a·b}{\,}^{\swarrow} &\hspace{12mm} {\,}^{\searrow}{\,}_{2·a·b}{\,}^{\swarrow} \\ \textbf{Step 2.} &\text{Write the square of the binomial.} &\quad &\hspace{13mm} (a+b)^2 &\hspace{13mm} (a−b)^2 \\ \textbf{Step 3.} &\text{Check by multiplying.} & & & \end{array}$$

This page titled 1.2.8: Factoring Special Products is shared under a CC BY license and was authored, remixed, and/or curated by OpenStax.