# 5.4 Special Products

## Learning Objectives

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

• Square a binomial using the Binomial Squares Pattern
• Multiply conjugates using the Product of Conjugates Pattern
• Recognize and use the appropriate special product pattern

### Try It

Before you get started, take this readiness quiz:

1) Simplify:
a. ${9}^{2}$
b. $(-9)^{2}$
c. $-{9}^{2}$

## Square a Binomial Using the Binomial Squares Pattern

Mathematicians like to look for patterns that will make their work easier. A good example of this is squaring binomials. While you can always get the product by writing the binomial twice and using the methods of the last section, there is less work to do if you learn to use a pattern.

 Let’s start by looking at $(x+9)^{2}$ What does this mean? $(x+9)^{2}$ It means to multiply $(x+9)$ by itself. $(x+9)(x+9)$ Then, using FOIL, we get: ${x}^{2}+9x+9x+81$ Combining like terms gives: ${x}^{2}+18x+81$
 Here’s another one: $(y-7)^{2}$ Multiply $(y-7)$ by itself. $(y-7)(y-7)$ Using FOIL, we get: $({y}^{2}-7y-7y+49$ And combining like terms: ${y}^{2}-14y+49$
 And one more: $(2x+3)^{2}$ Multiply. $(2x+3)(2x+3)$ Use FOIL: $4{x}^{2}+6x+6x+9$ Combine like terms. $4{x}^{2}+12x+9$

Look at these results. Do you see any patterns?

What about the number of terms? In each example, we squared a binomial and the result was a trinomial.

$(a+b)^{2}=$ ____ $+$ ____ $+$ ____

Now look at the first term in each result. Where did it come from?

$\begin{array}{ccc}({\color{red}{x}}+{\color{blue}{9}})^2\;\;\;&({\color{red}{y}}-{\color{blue}{7}})^2\;\;\;&({\color{red}{2x}}+{\color{blue}{3}})^2\\[2ex] ({\color{red}{x}}+{\color{blue}{9}})({\color{red}{x}}+{\color{blue}{9}})\;\;\;&({\color{red}{y}}-{\color{blue}{7}})({\color{red}{y}}-{\color{blue}{7}})\;\;\;&({\color{red}{2x}}+{\color{blue}{3}})({\color{red}{2x}}+{\color{blue}{3}})\\[2ex] {\color{red}{x^2}}+9x+9x+{\color{blue}{81}}\;\;\;&{\color{red}{y^2}}-7y-7y+{\color{blue}{49}}\;\;\;&{\color{red}{4x^2}}+6x+6x+{\color{blue}{9}}\\[2ex] {\color{red}{x^2}}+18x+{\color{blue}{81}}\;\;\;&{\color{red}{y^2}}-14y+{\color{blue}{49}}\;\;\;&{\color{red}{4x^2}}+12x+{\color{blue}{9}} \end{array}$

The first term is the product of the first terms of each binomial. Since the binomials are identical, it is just the square of the first term!

$(a+b)^{2}={a}^{2}+$ ____ $+$____

To get the first term of the product, square the first term.

Where did the last term come from? Look at the examples and find the pattern.

The last term is the product of the last terms, which is the square of the last term.

$(a+b)^{2}=$ ____ $+$ ____  $+{b}^{2}$

To get the last term of the product, square the last term.

Finally, look at the middle term. Notice it came from adding the “outer” and the “inner” terms—which are both the same! So the middle term is double the product of the two terms of the binomial.

$(a+b)^2=$  _____ $+ 2ab +$ ____

$(a-b)^2=$ ____ $- 2ab +$ _____

To get the middle term of the product, multiply the terms and double their product.

Putting it all together:

### Binomial Squares Pattern

if $a$ and $b$ are real numbers,

$(a+b)^2 = a^2 + 2ab + b^2$
$(a-b)^2 = a^2 - 2ab + b^2$
$\underbrace{(a+b)^2}_{(binomial)^2}\;\;=\;\;\underbrace{a^2\;}_{(first\;term)}\;+\;\underbrace{\;2ab}_{2(product\;of\;terms)}\;\;+\;\;\underbrace{b^2}_{(last\;term)^2}$

To square a binomial:

• square the first term
• square the last term
• double their product

A number example helps verify the pattern.

 $(10+4)^{2}$ Square the first term. ${10}^{2}+$____$+$ Square the last term. ${10}^{2}+$ ____$+{4}^{2}$ Double their product. ${10}^{2}+{2}\times{10}\times{4}+{4}^{2}$ Simplify. $100+80+16$ Simplify. $196$

To multiply $(10+4)^{2}$ usually you’d follow the Order of Operations.

$\begin{eqnarray*}&=&(10+4)^{2}\\&=&{14}^2\\&=&196\end{eqnarray*}$

The pattern works!

### Example 5.4.1

Multiply: $(x+5)^{2}$

Solution

$\left(\overset{{\color{red}{a\;+\;b}}}{x\;+\;5}\right)^2$

Step 1: Square the first term.

\begin{align*}&{\color{red}{a^2+2b+b^2}}\\&x^2+\underline{\;\;\;}+\underline{\;\;\;}\end{align*}

Step 2: Square the last term.

\begin{align*}&{\color{red}{a^2+2b+b^2}}\\ &x^2+\underline{\;\;\;}+5^2 \end{align*}

Step 3: Double the product.

\begin{align*}&{\color{red}{a^2+2\times a\times b+b^2}}\\&x^2+2\times x\times5+5^2\end{align*}

Step 4: Simplify.

$x^2\;+\;10x\;+\;25$

### Try It

2) Multiply: $(x+9)^{2}$

Solution

${x}^{2}+18x+81$

3) Multiply: $(y+11)^{2}$

Solution

${y}^{2}+22y+121$

### Example 5.4.2

Multiply: $(y-3)^{2}$

Solution

$\left(\overset{{\color{red}{a\;-\;b}}}{y\;-\;3}\right)^2$

Step 1: Square the first term.

\begin{align*}&{\color{red}{a^2-2b+b^2}}\\&y^2-\underline{\;\;\;}+\underline{\;\;\;}\end{align*}

Step 2: Square the last term.

\begin{align*}&{\color{red}{a^2-2b+b^2}}\\&y^2-\underline{\;\;\;}+3^2\end{align*}

Step 3: Double the product.

\begin{align*}&{\color{red}{a^2-2\times a\times b+b^2}}\\&y^2-2\times y\times3+3^2\end{align*}

Step 4: Simplify.

$y^2\;-\;6y\;+\;9$

### Try It

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

Solution

${x}^{2}-18x+81$

5) Multiply: $(p-13)^{2}$

Solution

${p}^{2}-26p+169$

### Example 5.4.3

Multiply: $(4x+6)^{2}$

Solution

$\left(\overset{{\color{red}{a\;\;\;\;+\;\;b}}}{4x\;+\;6}\right)^2$

Step 1: Use the pattern.

\begin{align*}&{\color{red}{\;a^2\;\;+2\times\;a\;\times b+b^2}}\\(&4x)^2+2\times 4x\times6+6^2\end{align*}

Step 2: Simplify.

$16x^2\;+\;48x\;+\;36$

### Try It

6) Multiply: $(6x+3)^{2}$

Solution

$36{x}^{2}+36x+9$

7) Multiply: $(4x+9)^{2}$

Solution

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

### Example 5.4.4

Multiply: $(2x-3y)^{2}$

Solution

$\left(\overset{{\color{red}{a\;-\;b}}}{2x\;-\;3y}\right)^2$

Step 1: Use the pattern.

\begin{align*}&{\color{red}{\;a^2\;\;-2\times\;a\;\times\;b\;+\;\;b^2}}\\(&2x)^2-2\times 2x\times3y+(3y)^2\end{align*}

Step 2: Simplify.

$4x^2\;-\;12xy\;+\;9y^2$

### Try It

8) Multiply: $(2c-d)^{2}$

Solution

$4{c}^{2}-4cd+{d}^{2}$

9) Multiply: $(4x-5y)^{2}$

Solution

$16{x}^{2}-40xy+25{y}^{2}$

### Example 5.4.5

Multiply: $(4{u}^{3}+1)^{2}$

Solution

$\left(\overset{{\color{red}{a\;\;\;\;+\;\;b}}}{4u^3\;+\;1}\right)^2$

Step 1: Use the pattern.

\begin{align*}&{\color{red}{\;a^2\;\;+2\times\;a\;\;\times b\;+\;b^2}}\\(&4u)^2+2\times 4u^3\times1+(1)^2\end{align*}

Step 2: Simplify.

$16u^6\;+\;8u^3\;+\;1$

### Try It

10) Multiply: $(2{x}^{2}+1)^{2}$

Solution

$4{x}^{4}+4{x}^{2}+1$

11) Multiply: $(3{y}^{3}+2)^{2}$

Solution

$9{y}^{6}+12{y}^{3}+4$

We just saw a pattern for squaring binomials that we can use to make multiplying some binomials easier. Similarly, there is a pattern for another product of binomials. But before we get to it, we need to introduce some vocabulary.

What do you notice about these pairs of binomials?

$(x-9)(x+9)$                       $(y-8)(y+8)$                  $(2x-5)(2x+5)$

Look at the first term of each binomial in each pair.

$({\color{red}{x}}-9)({\color{red}{x}}+9)$                       $({\color{red}{y}}-8)({\color{red}{y}}+8)$                 $({\color{red}{2x}}-5)({\color{red}{2x}}+5)$

Notice the first terms are the same in each pair.

Look at the last terms of each binomial in each pair.

$\underbrace{(a+b)^2}_{(binomial)^2}\;\;=\;\;\underbrace{a^2\;}_{(first\;term)}\;+\;\underbrace{\;2ab}_{2(product\;of\;terms)}\;\;+\;\;\underbrace{b^2}_{(last\;term)^2}$

Notice the last terms are the same in each pair.

Notice how each pair has one sum and one difference.

$\left(\underset{\overset{\color{blue}{\uparrow}}{Difference}}{x{\color{red}{-}}9}\right)\left(\underset{\overset{\color{blue}{\uparrow}}{Sum}}{x{\color{red}{+}}9}\right)\;\;\;\;\;\;\;\;\left(\underset{\overset{\color{blue}{\uparrow}}{Difference}}{y{\color{red}{-}}8}\right)\left(\underset{\overset{\color{blue}{\uparrow}}{Sum}}{y{\color{red}{+}}8}\right)\;\;\;\;\;\;\;\;\left(\underset{\overset{\color{blue}{\uparrow}}{Difference}}{2x{\color{red}{-}}5}\right)\left(\underset{\overset{\color{blue}{\uparrow}}{Sum}}{2x{\color{red}{+}}5}\right)$

A pair of binomials that each have the same first term and the same last term, but one is a sum and one is a difference has a special name. It is called a conjugate pair and is of the form $(a-b)$, $(a+b)$.

### Conjugate Pair

A conjugate pair is two binomials of the form

$(a-b)$, $(a+b)$

The pair of binomials each have the same first term and the same last term, but one binomial is a sum and the other is a difference.

There is a nice pattern for finding the product of conjugates. You could, of course, simply FOIL to get the product, but using the pattern makes your work easier.

Let’s look for the pattern by using FOIL to multiply some conjugate pairs.

 $(x-9)(x+9)$ $(y-8)(y+8)$ $(2x-5)(2x+5)$ $x^2+9x-9x-81$ $y^2+8y-8y-64$ $4x^2+10x-10x-25$ $x^2-81$ $y^2-64$ $4x^2-25$

$\begin{array}{ccc}({\color{red}{x}}{\color{blue}{+9}})({\color{red}{x}}{\color{blue}{-9}})\;\;\;&({\color{red}{y}}{\color{blue}{-8}})({\color{red}{y}}{\color{blue}{+8}})\;\;\;&({\color{red}{2x}}{\color{blue}{-5}})({\color{red}{2x}}{\color{blue}{+5}})\\[2ex]{\color{red}{x^2}}-9x+9x{\color{blue}{-81}}\;\;\;&{\color{red}{y^2}}+8y-8y{\color{blue}{-64}}\;\;\;&{\color{red}{4x^2}}+10x-10x{\color{blue}{-25}}\\[2ex] {\color{red}{x^2}}{\color{blue}{-81}}\;\;\;&{\color{red}{y^2}}{\color{blue}{-64}}\;\;\;&{\color{red}{4x^2}}{\color{blue}{-25}}\end{array}$

Each first term is the product of the first terms of the binomials, and since they are identical it is the square of the first term.

$(a+b)(a-b)=a^2$____
To get the first term, square the first term.

The last term came from multiplying the last terms, the square of the last term.

$(a+b)(a-b)=a^2-b^2$
To get the last term, square the last term.

#### What do you observe about the products?

The product of the two binomials is also a binomial! Most of the products resulting from FOIL have been trinomials.

Why is there no middle term? Notice the two middle terms you get from FOIL combine to 0 in every case, the result of one addition and one subtraction.

The product of conjugates is always of the form ${a}^{2}-{b}^{2}$ . This is called a difference of squares.

### Product of Conjugates Pattern

If $a$ and $b$ are real numbers,

The product is called a difference of squares.

To multiply conjugates, square the first term, square the last term, and write the product as a difference of squares.

Let’s test this pattern with a numerical example.

 $(10-2)(10+2)$ It is the product of conjugates, so the result will be the difference of two squares. ____ $-$ ____ Square the first term. $10^2-$____ Square the last term. $10^2-2^2$ Simplify. $100-4\\=96$ What do you get using the order of operations? $(10 - 2)(10 + 2)\\(8)(12)\\=96$

Notice, the result is the same!

### Example 5.4.6

Multiply: $(x - 8)(x + 8)$

Solution

First, recognize this as a product of conjugates. The binomials have the same first terms, and the same last terms, and one binomial is a sum and the other is a difference.

Step 1: It fits the pattern.

$\left(\overset{{\color{red}{a\;-\;b}}}{x\;-\;8}\right)\left(\overset{{\color{red}{a\;+\;b}}}{x\;+\;8}\right)$

Step 2: Square the first term, $x$.

$\overset{{\color{red}{a^2\;-\;b^2}}}{x^2\;-\;\_\_}$

Step 3: Square the last term, $8$.

$\overset{{\color{red}{a^2\;-\;b^2}}}{x^2\;-\;8^2}$

Step 4: The product is a difference of squares.

$\overset{{\color{red}{a^2\;-\;b^2}}}{x^2\;-\;64}$

### Try It

12) Multiply: $(x-5)(x+5)$

Solution

${x}^{2}-25$

13) Multiply: $(w-3)(w+3)$

Solution

${w}^{2}-9$

### Example 5.4.7

Multiply: $(2x+5)(2x-5)$

Solution

Are the binomials conjugates?

Step 1: It is the product of conjugates.

$\left(\overset{{\color{red}{a\;+\;b}}}{2x\;+\;5}\right)\left(\overset{{\color{red}{a\;-\;b}}}{2x\;-\;5}\right)$

Step 2: Square the first term, $2x$.

$\overset{{\color{red}{a^2\;-\;b^2}}}{{(2x)}^2\;-\;\_\_}$

Step 3: Square the last term, $5$.

$\overset{{\color{red}{a^2\;-\;b^2}}}{{(2x)}^2\;-\;5^2}$

Step 4: Simplify. The product is a difference of squares.

$\overset{{\color{red}{a^2\;-\;b^2}}}{4x^2\;-\;25}$

### Try It

14) Multiply: $(6x+5)(6x-5)$.

Solution

$36{x}^{2}-25$

15) Multiply: $(2x+7)(2x-7)$

Solution

$4{x}^{2}-49$

The binomials in the next example may look backwards – the variable is in the second term. But the two binomials are still conjugates, so we use the same pattern to multiply them.

### Example 5.4.8

Find the product: $(3+5x)(3-5x)$

Solution

Step 1: It is the product of conjugates.

$\left(\overset{{\color{red}{a\;-\;b}}}{3\;+\;5x}\right)\left(\overset{{\color{red}{a\;+\;b}}}{3\;-\;5x}\right)$

Step 2: Use the pattern.

$\overset{{\color{red}{a^2\;-\;b^2}}}{3^2\;-\;{(5x)}^2}$

Step 3: Simplify.

$9\;-\;25x^2$

### Try It

16) Multiply: $(7+4x)(7-4x)$

Solution

$49-16{x}^{2}$

17) Multiply: $(9-2y)(9+2y)$

Solution

$81-4y^{2}$

Now we’ll multiply conjugates that have two variables.

### Example 5.4.9

Find the product: $(5m-9n)(5m+9n)$

Solution

Step 1: This fits the pattern.

$\left(\overset{{\color{red}{a\;-\;b}}}{5m\;-\;9n}\right)\left(\overset{{\color{red}{a\;+\;b}}}{5m\;+\;9n}\right)$

Step 2: Use the pattern.

$\overset{{\color{red}{a^2\;-\;b^2}}}{{(5m)}^2\;-\;{(9n)}^2}$

Step 3: Simplify.

$25m^2\;-\;81n^2$

### Try It

18) Find the product: $(4p-7q)(4p+7q)$

Solution

$16{p}^{2}-49{q}^{2}$

19) Find the product: $(3x-y)(3x+y)$

Solution

$9{x}^{2}-{y}^{2}$

### Example 5.4.10

Find the product: $(cd-8)(cd+8)$

Solution

Step 1: This fits the pattern.

$\left(\overset{{\color{red}{a\;-\;b}}}{cd\;-\;8}\right)\left(\overset{{\color{red}{a\;+\;b}}}{cd\;+\;8}\right)$

Step 2: Use the pattern.

$\overset{{\color{red}{a^2\;-\;b^2}}}{{(cd)}^2\;-\;{(8)}^2}$

Step 3: Simplify.

$c^2d^2\;-\;64$

### Try It

20) Find the product: $(xy-6)(xy+6)$

Solution

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

21) Find the product: $(ab-9)(ab+9)$

Solution

${a}^{2}{b}^{2}-81$

### Example 5.4.11

Find the product: $(6{u}^{2}-11{v}^{5})(6{u}^{2}+11{v}^{5})$

Solution

Step 1: This fits the pattern.

$\left(\overset{{\color{red}{a\;-\;b}}}{6u^2\;-\;11v^5}\right)\left(\overset{{\color{red}{a\;+\;b}}}{6u^2\;+\;11v^5}\right)$

Step 2: Use the pattern.

$\overset{{\color{red}{a^2\;-\;b^2}}}{\left(6u^2\right)^2-\;\left(11v^5\right)^2}$

Step 3: Simplify.

$36u^4\;-\;121v^{10}$

### Try It

22) Find the product: $(3{x}^{2}-4{y}^{3})(3{x}^{2}+4{y}^{3})$

Solution

$9{x}^{4}-16{y}^{6}$

23) Find the product: $(2{m}^{2}-5{n}^{3})(2{m}^{2}+5{n}^{3})$

Solution

$4{m}^{4}-25{n}^{6}$

## Recognize and Use the Appropriate Special Product Pattern

We just developed special product patterns for Binomial Squares and for the Product of Conjugates. The products look similar, so it is important to recognize when it is appropriate to use each of these patterns and to notice how they differ. Look at the two patterns together and note their similarities and differences.

### Comparing the Special Product Patterns

$(a-b)(a+b)={a}^{2}-{b}^{2}$

Binomial Squares Product of Conjugates
$(a+b)^{2}={a}^{2}+2ab+{b}^{2}$ ${(a-b)}^{2}={a}^{2}-2ab+{b}^{2}$
– Squaring a binomial – Multiplying conjugates
– Product is a trinomial – Product is a binomial
– Inner and outer terms with FOIL are the same. – Inner and outer terms with FOIL are opposites.
– Middle term is double the product of the terms. – There is no middle term.

### Example 5.4.12

Choose the appropriate pattern and use it to find the product:

a. $(2x-3)(2x+3)$
b. $(8x-5)^{2}$
c. $(6m+7)^{2}$
d. $(5x-6)(6x+5)$

Solution

a. $(2x-3)(2x+3)$ These are conjugates. They have the same first numbers, and the same last numbers, and one binomial is a sum and the other is a difference. It fits the Product of Conjugates pattern.

Step 1: This fits the pattern.

$\left(\overset{{\color{red}{a\;-\;b}}}{2x\;-\;3}\right)\left(\overset{{\color{red}{a\;+\;b}}}{2x\;+\;3}\right)$

Step 2: Use the pattern.

\begin{align*}&{\color{red}{\;a^2\;\;\;-b^2}}\\(&2x)^2-3^2\\[2ex]\text{Simplify.}\;\;&4x^2\;-\;9\end{align*}

b. $(8x-5)^{2}$ We are asked to square a binomial. It fits the binomial squares pattern.

$\left(\overset{{\color{red}{a\;-\;b}}}{8x\;-\;5}\right)^2$

Step 1: Use the pattern.

\begin{align*} &\overset{{\color{red}{a^2\;-\;2ab\;+\;b^2}}}{{(8x)}^2\;-\;2\;\times\;8x\;\times\;5\;+\;5^2}\\ \text{Simplify.}\;\;&64x^2\;-\;80x\;+\;25 \end{align*}

c. $(6m+7)^{2}$ Again, we will square a binomial so we use the binomial squares pattern.

$\left(\overset{{\color{red}{a\;+\;b}}}{6m\;+\;7}\right)^2$

Step 1: Use the pattern.

\begin{align*}&\overset{{\color{red}{a^2\;+\;2ab\;+\;b^2}}}{\left(6m\right)^2\;+\;2\;\times\;6m\;\times\;7\;+7^2}\\\text{Simplify.}\;\;&36m^2\;+\;84m\;+\;49\end{align*}

d. $(5x-6)(6x+5)$ This product does not fit the patterns, so we will use FOIL.

Step 1: Use FOIL.

\begin{align*}&30{x}^{2}+25x-36x-30\\\text{Simplify.}\;\;&30{x}^{2}-11x-30\end{align*}

### Try It

24) Choose the appropriate pattern and use it to find the product:

a. $(9b-2)(2b+9)$
b. $(9p-4)^{2}$
c. $(7y+1)^{2}$
d. $(4r-3)(4r+3)$

Solution

a. FOIL;  $18{b}^{2}+77b-18$
b. Binomial Squares; $81{p}^{2}-72p+16$
c. Binomial Squares; $49{y}^{2}+14y+1$
d. Product of Conjugates; $16{r}^{2}-9$

25) Choose the appropriate pattern and use it to find the product:

a. $(6x+7)^{2}$
b. $(3x-4)(3x+4)$
c. $(2x-5)(5x-2)$
d. $(6n-1)^{2}$

Solution

a. Binomial Squares; $36{x}^{2}+84x+49$
b. Product of Conjugates; $9{x}^{2}-16$
c. FOIL; $10{x}^{2}-29x+10$
d. Binomial Squares;  $36{n}^{2}-12n+1$

Access these online resources for additional instruction and practice with special products:

### Key Concepts

• Binomial Squares Pattern
• If $a$ and $b$ are real numbers,
${\underbrace{{(a\;+\;b)}^2}_{\left(binomial\right)^2}\;}=\;\underbrace{a^2}_{\left(first\;term\right)^2}\;+\;\underbrace{2ab}_{2(product\;of\;terms)}\;+\;\;\underbrace{b^2}_{{(last\;term)}^2}$
• $(a+b)^{2}={a}^{2}+2ab+{b}^{2}$
• $(a-b)^{2}={a}^{2}-2ab+{b}^{2}$
• To square a binomial: square the first term, square the last term, double their product.

• Product of Conjugates Pattern
• If $a$ and $b$ are real numbers
$\underbrace{(a\;-\;b)(a\;+\;b)}_{conjugates}\;=\;\underset{squares}{\underbrace{a^{2\;}}\;\overset{\overset{difference}\downarrow}-\;\underbrace{b^2}}$
• $(a-b)(a+b)={a}^{2}-{b}^{2}$
• The product is called a difference of squares.

• To multiply conjugates:
• square the first term square the last term write it as a difference of squares

### Self Check

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

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?

### Glossary

conjugate pair
A conjugate pair is two binomials of the form $(a-b)$, $(a+b)$; the pair of binomials each have the same first term and the same last term, but one binomial is a sum and the other is a difference.
definition