6.3 Special Products

Pooja Gupta

6.3 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

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 ${\left(x+9\right)}^{2}$ .
What does this mean?	${\left(x+9\right)}^{2}$
It means to multiply $\left(x+9\right)$ by itself.	$\left(x+9\right)\left(x+9\right)$
Then, using FOIL, we get:	${x}^{2}+9x+9x+81$
Combining like terms gives:	${x}^{2}+18x+81$

Here’s another one:	${\left(y-7\right)}^{2}$
Multiply $\left(y-7\right)$ by itself.	$\left(y-7\right)\left(y-7\right)$
Using FOIL, we get:	${y}^{2}-7y-7y+49$
And combining like terms:	${y}^{2}-14y+49$

And one more:	${\left(2x+3\right)}^{2}$
Multiply.	$\left(2x+3\right)\left(2x+3\right)$
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.

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

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

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!

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

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.

${\left(a+b\right)}^{2}=\underline{\qquad}+\underline{\qquad}+{b}^{2}$

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.

${\left(a+b\right)}^{2}=\underline{\qquad}+ 2ab+ \underline{\qquad}$

${\left(a-b\right)}^{2}=\underline{\qquad}-2ab + \underline{\qquad}$

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,

$\begin{array}{c} {\left(a+b\right)}^{2}={a}^{2}+2ab+{b}^{2}\hfill \\ {\left(a-b\right)}^{2}={a}^{2}-2ab+{b}^{2}\hfill \end{array}$

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HOW TO:

To square a binomial:

square the first term
square the last term
double their product

A number example helps verify the pattern.

${\left(10+4\right)}^{2}$

Square the first term.

${10}^{2}+\underline{\qquad}+$

Square the last term.

${10}^{2}+\underline{\qquad}+{4}^{2}$

Double their product.

${10}^{2}+2\cdot 10 \cdot 4+{4}^{2}$

Simplify.

$100+80+16$

Simplify.

$196$

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

$\begin{array}{c} {\left(10+4\right)}^{2} \\ {\left(14\right)}^{2} \\ 196 \end{array}$

The pattern works!

EXAMPLE 1

Multiply: ${\left(x+5\right)}^{2}$ .

Solution

	$\left(\begin{array}{c} \color{red}a+b \\ x+5 \end{array}\right)^2$
Square the first term.	$\begin{aligned} & \color{red}a^2+2 a b+b^2 \\ & x^2+\rule{0.7cm}{0.4pt}+\rule{0.7cm}{0.4pt} \\ \end{aligned}$
Square the last term.	$\begin{aligned} & \color{red}a^2+2 a b+b^2 \\ & x^2+\rule{0.7cm}{0.4pt}+5^2 \end{aligned}$
Double the product.	$\begin{aligned} & \color{red}a^2+2 \cdot a \cdot b+b^2 \\ & x^2+2 \cdot x \cdot 5+25 \end{aligned}$
Simplify.	$x^2+10x+25$

TRY IT 1.1

Multiply: ${\left(x+9\right)}^{2}$ .

Show answer

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

TRY IT 1.2

Multiply: ${\left(y+11\right)}^{2}$ .

Show answer

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

EXAMPLE 2

Multiply: ${\left(y-3\right)}^{2}$ .

Solution

	$\left(\begin{array}{c} \color{red}a-b \\ y-3 \end{array}\right)^2$
Square the first term.	$\begin{aligned} & \color{red}a^2-2 a b+b^2 \\ & y^2-\rule{0.7cm}{0.4pt}+\rule{0.7cm}{0.4pt} \\ \end{aligned}$
Square the last term.	$\begin{aligned} & \color{red}a^2-2 a b+b^2 \\ & y^2-\rule{0.7cm}{0.4pt}+3^2 \end{aligned}$
Double the product.	$\begin{aligned} & \color{red}a^2-2 \cdot a \cdot b+b^2 \\ & y^2-2 \cdot y \cdot 3+9 \end{aligned}$
Simplify.	$y^2-6y+9$

TRY IT 2.1

Multiply: ${\left(x-9\right)}^{2}$ .

Show answer

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

TRY IT 2.2

Multiply: ${\left(p-13\right)}^{2}$ .

Show answer

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

EXAMPLE 3

Multiply: ${\left(4x+6\right)}^{2}$ .

Solution

	$\left(\begin{array}{c} \color{red}a+b \\ 4x+6 \end{array}\right)^2$
Square the first term.	$\begin{aligned} & \color{red}a^2+2 a b+b^2 \\ & (4x)^2+\rule{0.7cm}{0.4pt}+\rule{0.7cm}{0.4pt} \\ \end{aligned}$
Square the last term.	$\begin{aligned} & \color{red}a^2+2 a b+b^2 \\ & 16x^2+\rule{0.7cm}{0.4pt}+6^2 \end{aligned}$
Double the product.	$\begin{aligned} & \color{red}a^2+2 \cdot a \cdot b+b^2 \\ & 16x^2+2 \cdot 4x \cdot 6+36 \end{aligned}$
Simplify.	$16x^2+48x+36$

Once you become comfortable with using the pattern (or formula), you can apply it after determining the values of $a$ and $b$ .

$\left(\begin{array}{c} \color{red}a+b \\ 4x+6 \end{array}\right)^2$

Here the value of $a$ is $4x$ and $b$ is $6$ . That means you can substitute $a=4x$ and $b=6$ in the pattern for squaring a binomial.

Use the pattern or formula.

$\begin{centered} & {\color{red}a^2+2 \cdot a \cdot b+b^2} \\ & (4x)^2+2 \cdot 4x \cdot 6+ (6)^2 \end{centered}$

Simplify.

$16x^2+48x+36$

TRY IT 3.1

Multiply: ${\left(6x+3\right)}^{2}$ .

Show answer

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

TRY IT 3.2

Multiply: ${\left(4x+9\right)}^{2}$ .

Show answer

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

EXAMPLE 4

Multiply: ${\left(2x-3y\right)}^{2}$ .

Solution

$\left(\begin{array}{c} \color{red}a-b \\ 2x-3y \end{array}\right)^2$

Here the value of $a$ is $2x$ and $b$ is $3y$ . That means you can substitute $a=2x$ and $b=3y$ in the pattern for squaring a binomial.

Use the pattern.

$\begin{centered} & {\color{red}a^2-2 \cdot a \cdot b+b^2} \\ & (2x)^2-2 \cdot 2x \cdot 3y+ (3y)^2 \end{centered}$

Simplify.

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

TRY IT 4.1

Multiply: ${\left(2c-d\right)}^{2}$ .

Show answer

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

TRY IT 4.2

Multiply: ${\left(4x-5y\right)}^{2}$ .

Show answer

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

EXAMPLE 5

Multiply: ${\left(4{u}^{3}+1\right)}^{2}$ .

Solution

$\left(\begin{array}{c} \color{red}a+b \\ 4u^3+1 \end{array}\right)^2$

Here the value of $a$ is $4u^3$ and $b$ is $1$ . That means you can substitute $a=4u^3$ and $b=1$ in the pattern for squaring a binomial.

Use the pattern or formula.

$\begin{centered} & {\color{red}a^2 + \quad 2 \cdot a \cdot b+ \quad b^2} \\ & (4u^3)^2+2 \cdot 4u^3 \cdot 1+ (1)^2 \end{centered}$

Simplify.

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

TRY IT 5.1

Multiply: ${\left(2{x}^{2}+1\right)}^{2}$ .

Show answer

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

TRY IT 5.2

Multiply: ${\left(3{y}^{3}+2\right)}^{2}$ .

Show answer

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

Multiply Conjugates Using the Product of Conjugates Pattern

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?

$\left(x-9\right)\left(x+9\right)\qquad \qquad \left(y-8\right)\left(y+8\right)\qquad \qquad\left(2x-5\right)\left(2x+5\right)$

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

This figure has three products. The first is x minus 9, in parentheses, times x plus 9, in parentheses. The second is y minus 8, in parentheses, times y plus 8, in parentheses. The last is 2x minus 5, in parentheses, times 2x plus 5, in parentheses

Notice the first terms are the same in each pair.

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

This figure has three products. The first is x minus 9, in parentheses, times x plus 9, in parentheses. The second is y minus 8, in parentheses, times y plus 8, in parentheses. The last is 2x minus 5, in parentheses, times 2x plus 5, in parentheses.

Notice the last terms are the same in each pair.

Notice how each pair has one sum and one difference.

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 $\left(a-b\right),\left(a+b\right)$ .

Conjugate Pair

A conjugate pair is two binomials of the form

$\left(a-b\right),\left(a+b\right)$ .

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.

$\begin{array}{ccc} \left(x-9\right)\left(x+9\right) & \left(y-8\right)\left(y+8\right) & \left(2x-5\right)\left(2x+5\right) \\ {x}^{2}+9x-9x-81 \qquad &{y}^{2}+8y-8y-64 \qquad & 4{x}^{2}+10x-10x-25 \\ {x}^{2}-81 & {y}^{2}-64 & 4{x}^{2}-25 \end{array}$

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.

$\begin{array}{c} \left(a+b\right)\left(a-b\right)={a}^{2}-\underline{\qquad} \\ \text{To get the}\textbf{first term, square the first term}. \end{array}$

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

$\begin{array}{c} \left(a+b\right)\left(a-b\right)={a}^{2}-{b}^{2} \\ \text{To get the}\textbf{last term, square the last term}. \end{array}$

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.

This leads to the pattern:

Product of Conjugates Pattern

If $a$ and $b$ are real numbers,
$(a+b)(a-b)= a^2- b^2$

image showing product of two conjugates

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.

	$\left(10-2\right)\left(10+2\right)$
It is the product of conjudgates, so the result will be the difference of two squares.	____ – ____
Square the first term.	${10}^{2}-\underline{\qquad}$
Square the last term.	${10}^{2}-{2}^{2}$
Simplify.	$100-4$
Simplify.	$96$
What do you get using the order of operations?
	$\begin{array}{c}\left(10-2\right)\left(10+2\right)\\ \left(8\right)\left(12\right)\\ 96\end{array}$

Notice, the result is the same!

EXAMPLE 6

Multiply: $\left(x-8\right)\left(x+8\right)$ .

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.

It fits the pattern.	$\left(\begin{array}{l} \color{red}a-b \\ x-8 \end{array}\right)\left(\begin{array}{l} \color{red}a+b \\ x+8 \end{array}\right)$
Square the first term, x.	$\begin{aligned} & \color{red}a^2 - b^2 \\ & x^2 - \rule{0.5cm}{0.4pt} \\ \end{aligned}$
Square the last term, 8.	$\begin{aligned} & \color{red}a^2 - b^2 \\ & x^2 - 8^2 \\ \end{aligned}$
The product is a difference of squares.	$x^2-64$

TRY IT 6.1

Multiply: $\left(x-5\right)\left(x+5\right)$ .

Show answer

${x}^{2}-25$

TRY IT 6.2

Multiply: $\left(w-3\right)\left(w+3\right)$ .

Show answer

${w}^{2}-9$

EXAMPLE 7

Multiply: $\left(2x+5\right)\left(2x-5\right)$ .

Solution

Are the binomials conjugates? YES

It is the product of conjugates.	$\left(\begin{array}{l} \color{red}a+b \\ 2x+5 \end{array}\right)\left(\begin{array}{l} \color{red}a-b \\ 2x-5 \end{array}\right)$
Square the first term, $2x$ .	$\begin{centered} & {\color{red}a^2 - b^2} \\ & (2x)^2 - \rule{0.5cm}{0.4pt} \end{centered}$
Square the last term, $5$ .	$\begin{aligned} & {\color{red}a^2 - b^2} \\ & 4x^2 - 5^2 \end{aligned}$
Simplify. The product is a difference of squares.	$4x^2-25$

TRY IT 7.1

Multiply: $\left(6x+5\right)\left(6x-5\right)$ .

Show answer

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

TRY IT 7.2

Multiply: $\left(2x+7\right)\left(2x-7\right)$ .

Show answer

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

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 8

Find the product: $\left(3+5x\right)\left(3-5x\right)$ .

Solution

It is the product of conjugates.	$\left(\begin{array}{l} \color{red}a+b \\ 3+5x \end{array}\right)\left(\begin{array}{l} \color{red}a-b \\ 3-5x \end{array}\right)$
Use the pattern. Here $a=3$ and $b=5x$ .	$\begin{aligned} & {\color{red}a^2 - b^2} \\ & 3^2 - (5x)^2 \end{aligned}$
Simplify.	$9-25x^2$

TRY IT 8.1

Multiply: $\left(7+4x\right)\left(7-4x\right)$ .

Show answer

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

TRY IT 8.2

Multiply: $\left(9-2y\right)\left(9+2y\right)$ .

Show answer

$81-4{y}^{2}$

Now we’ll multiply conjugates that have two variables.

EXAMPLE 9

Find the product: $\left(5m-9n\right)\left(5m+9n\right)$ .

Solution

This fits the pattern of product of conjugates.	$\left(\begin{array}{l} \color{red}\quad a-b \\ 5m-9n \end{array}\right)\left(\begin{array}{l} \color{red} \quad a+b \\ 5m+9n \end{array}\right)$
Use the pattern. Here $a = 5m$ and $b = 9n$ .	$\begin{aligned} & {\color{red} \quad a^2 - b^2} \\ & (5m)^2 - (9n)^2 \end{aligned}$
Simplify.	$25m^2-81n^2$

TRY IT 9.1

Find the product: $\left(4p-7q\right)\left(4p+7q\right)$ .

Show answer

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

TRY IT 9.2

Find the product: $\left(3x-y\right)\left(3x+y\right)$ .

Show answer

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

EXAMPLE 10

Find the product: $\left(cd-8\right)\left(cd+8\right)$ .

Solution

This fits the pattern of product of conjugates.	$\left(\begin{array}{l} \color{red}\hspace{0.2cm} a-b \\ cd-8 \end{array}\right)\left(\begin{array}{l} \color{red} \hspace{0.2cm} a+b \\ cd+8 \end{array}\right)$
Use the pattern. Here $a = cd$ and $b = 8$ .	$\begin{aligned} & {\color{red} \quad a^2 - b^2} \\ & (cd)^2 - (8)^2 \end{aligned}$
Simplify.	$c^2d^2-64$

TRY IT 10.1

Find the product: $\left(xy-6\right)\left(xy+6\right)$ .

Show answer

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

TRY IT 10.2

Find the product: $\left(ab-9\right)\left(ab+9\right)$ .

Show answer

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

EXAMPLE 11

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

Solution

This fits the pattern of product of conjugates.	$\left(\begin{array}{l} \color{red}\hspace{0.2cm} a \hspace{0.2cm}- \hspace{0.2cm}b \\ 6u^2-11v^5 \end{array}\right)\left(\begin{array}{l} \color{red} \hspace{0.2cm} a \hspace{0.2cm}+ \hspace{0.2cm}b \\ 6u^2+11v^5 \end{array}\right)$
Use the pattern. Here $a = 6u^2$ and $b = 11v^5$ .	$\begin{aligned} & {\color{red} \quad a^2 \hspace{0.2cm} - \hspace{0.3cm} b^2} \\ & (6u^2)^2 - (11v^5)^2 \end{aligned}$
Simplify.	$36^4-121v^{10}$

TRY IT 11.1

Find the product: $\left(3{x}^{2}-4{y}^{3}\right)\left(3{x}^{2}+4{y}^{3}\right)$ .

Show answer

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

TRY IT 11.2

Find the product: $\left(2{m}^{2}-5{n}^{3}\right)\left(2{m}^{2}+5{n}^{3}\right)$ .

Show answer

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

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

Binomial Squares

Product of Conjugates

${\left(a+b\right)}^{2}={a}^{2}+2ab+{b}^{2}$

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

${\left(a-b\right)}^{2}={a}^{2}-2ab+{b}^{2}$

– 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 12

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

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

Solution

$\left(2x-3\right)\left(2x+3\right)$ 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.

$\left(\begin{array}{l} \color{red}\hspace{0.2cm} a-b \\ 2x-3 \end{array}\right)\left(\begin{array}{l} \color{red} \hspace{0.2cm} a+b \\ 2x+3 \end{array}\right)$

Use the pattern. $\begin{aligned} & {\color{red} \quad a^2 - b^2} \\ & (2x)^2 - (3)^2 \end{aligned}$

Simplify. $4x^2-9$
${\left(8x-5\right)}^{2}$ We are asked to square a binomial. It fits the binomial squares pattern.

$\left(\begin{array}{c} \color{red}a-b \\ 8x-5 \end{array}\right)^2$

Use the pattern. $\begin{centered} & {\color{red}a^2-2 \cdot a \cdot b+b^2} \\ & (8x)^2-2 \cdot 8x \cdot 5+ (5)^2 \end{centered}$

Simplify. $64x^2-80x+25$
${\left(6m+7\right)}^{2}$ Again, we will square a binomial so we use the binomial squares pattern.

$\left(\begin{array}{c} \color{red}a+b \\ 6m+7 \end{array}\right)^2$

Use the pattern. $\begin{centered} & {\color{red}a^2+2 \cdot a \cdot b+b^2} \\ & (6m)^2+2 \cdot 6m \cdot 7+ (7)^2 \end{centered}$

Simplify. $36m^2+84m+49$
$\left(5x-6\right)\left(6x+5\right)$ This product does not fit the patterns, so we will use FOIL.

$\left(5x-6\right)\left(6x+5\right)$

Use FOIL. $30{x}^{2}+25x-36x-30$

Simplify. $30{x}^{2}-11x-30$

TRY IT 12.1

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

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

Show answer

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; $16{r}^{2}-9$

TRY IT 12.2

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

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

Show answer

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:

Special Products

Key Concepts

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

Glossary

conjugate pair: A conjugate pair is two binomials of the form $\left(a-b\right),\left(a+b\right)$ ; 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.

Practice Makes Perfect

Square a Binomial Using the Binomial Squares Pattern

In the following exercises, square each binomial using the Binomial Squares Pattern.

1. ${\left(q+12\right)}^{2}$	2. ${\left(w+4\right)}^{2}$
3. ${\left(x+\dfrac{2}{3}\right)}^{2}$	4. ${\left(y+\dfrac{1}{4}\right)}^{2}$
5. ${\left(y-6\right)}^{2}$	6. ${\left(b-7\right)}^{2}$
7. ${\left(p-13\right)}^{2}$	8. ${\left(m-15\right)}^{2}$
9. ${\left(4a+10\right)}^{2}$	10. ${\left(3d+1\right)}^{2}$
11. ${\left(3z+\dfrac{1}{5}\right)}^{2}$	12. ${\left(2q+\dfrac{1}{3}\right)}^{2}$
13. ${\left(2y-3z\right)}^{2}$	14. ${\left(3x-y\right)}^{2}$
15. ${\left(\dfrac{1}{8}x-\dfrac{1}{9}y\right)}^{2}$	16. ${\left(\dfrac{1}{5}x-\dfrac{1}{7}y\right)}^{2}$
17. ${\left(5{u}^{2}+9\right)}^{2}$	18. ${\left(3{x}^{2}+2\right)}^{2}$
19. ${\left(8{p}^{3}-3\right)}^{2}$	20. ${\left(4{y}^{3}-2\right)}^{2}$

In the following exercises, multiply each pair of conjugates using the Product of Conjugates Pattern.

Multiply Conjugates Using the Product of Conjugates Pattern

21. $\left(c-5\right)\left(c+5\right)$	22. $\left(m-7\right)\left(m+7\right)$
23. $\left(b+\dfrac{6}{7}\right)\left(b-\dfrac{6}{7}\right)$	24. $\left(x+\dfrac{3}{4}\right)\left(x-\dfrac{3}{4}\right)$
25. $\left(8j+4\right)\left(8j-4\right)$	26. $\left(5k+6\right)\left(5k-6\right)$
27. $\left(9c+5\right)\left(9c-5\right)$	28. $\left(11k+4\right)\left(11k-4\right)$
29. $\left(13-q\right)\left(13+q\right)$	30. $\left(11-b\right)\left(11+b\right)$
31. $\left(4-6y\right)\left(4+6y\right)$	32. $\left(5-3x\right)\left(5+3x\right)$
33. $\left(7w+10x\right)\left(7w-10x\right)$	34. $\left(9c-2d\right)\left(9c+2d\right)$
35. $\left(p+\dfrac{4}{5}q\right)\left(p-\dfrac{4}{5}q\right)$	36. $\left(m+\dfrac{2}{3}n\right)\left(m-\dfrac{2}{3}n\right)$
37. $\left(xy-9\right)\left(xy+9\right)$	38. $\left(ab-4\right)\left(ab+4\right)$
39. $\left(rs-\dfrac{2}{7}\right)\left(rs+\dfrac{2}{7}\right)$	40. $\left(uv-\dfrac{3}{5}\right)\left(uv+\dfrac{3}{5}\right)$
41. $\left(6{m}^{3}-4{n}^{5}\right)\left(6{m}^{3}+4{n}^{5}\right)$	42. $\left(2{x}^{2}-3{y}^{4}\right)\left(2{x}^{2}+3{y}^{4}\right)$
43. $\left(15{m}^{2}-8{n}^{4}\right)\left(15{m}^{2}+8{n}^{4}\right)$	44. $\left(12{p}^{3}-11{q}^{2}\right)\left(12{p}^{3}+11{q}^{2}\right)$

In the following exercises, find each product.

Recognize and Use the Appropriate Special Product Pattern

45.

a) ${\left(2r+12\right)}^{2}$

b) $\left(3p+8\right)\left(3p-8\right)$

c) $\left(7a+b\right)\left(a-7b\right)$

d) ${\left(k-6\right)}^{2}$

46.

a) $\left(p-3\right)\left(p+3\right)$

b) ${\left(t-9\right)}^{2}$

c) ${\left(m+n\right)}^{2}$

d) $\left(2x+y\right)\left(x-2y\right)$

47.

a) $\left({x}^{5}+{y}^{5}\right)\left({x}^{5}-{y}^{5}\right)$

b) ${\left({m}^{3}-8n\right)}^{2}$

c) ${\left(9p+8q\right)}^{2}$

d) $\left({r}^{2}-{s}^{3}\right)\left({r}^{3}+{s}^{2}\right)$

48.

a) ${\left({a}^{5}-7b\right)}^{2}$

b) $\left({x}^{2}+8y\right)\left(8x-{y}^{2}\right)$

c) $\left({r}^{6}+{s}^{6}\right)\left({r}^{6}-{s}^{6}\right)$

d) ${\left({y}^{4}+2z\right)}^{2}$

Everyday Math

49. Mental math You can use the binomial squares pattern to multiply numbers without a calculator. Say you need to square 65. Think of 65 as $60+5$ .

Multiply ${\left(60+5\right)}^{2}$ by using the binomial squares pattern, ${\left(a+b\right)}^{2}={a}^{2}+2ab+{b}^{2}$ .
Square 65 without using a calculator.
Which way is easier for you? Why?

50. Mental math You can use the product of conjugates pattern to multiply numbers without a calculator. Say you need to multiply 47 times 53. Think of 47 as $50-3$ and 53 as $50+3$ .

Multiply $\left(50-3\right)\left(50+3\right)$ by using the product of conjugates pattern, $\left(a-b\right)\left(a+b\right)={a}^{2}-{b}^{2}$ .
Multiply $47\cdot 53$ without using a calculator.
Which way is easier for you? Why?

Writing Exercises

52. Why does ${\left(a+b\right)}^{2}$ result in a trinomial, but $\left(a-b\right)\left(a+b\right)$ result in a binomial?	51. How do you decide which pattern to use?
54. Use the order of operations to show that ${\left(3+5\right)}^{2}$ is 64, and then use that numerical example to explain why ${\left(a+b\right)}^{2}\ne {a}^{2}+{b}^{2}$ .	53. Marta did the following work on her homework paper: $\begin{array}{c}{\left(3-y\right)}^{2} \\{3}^{2}-{y}^{2} \\9-{y}^{2} \end{array}$ Explain what is wrong with Marta’s work.

Answers

1. ${q}^{2}+24q+144$	3. ${x}^{2}+\dfrac{4}{3}x+\dfrac{4}{9}$
5. ${y}^{2}-12y+36$	7. ${p}^{2}-26p+169$
9. $16{a}^{2}+80a+100$	11. $9{z}^{2}+\dfrac{6}{5}z+\dfrac{1}{25}$
13. $4{y}^{2}-12yz+9{z}^{2}$	15. $\dfrac{1}{64}{x}^{2}-\dfrac{1}{36}xy+\dfrac{1}{81}{y}^{2}$
17. $25{u}^{4}+90{u}^{2}+81$	19. $64{p}^{6}-48{p}^{3}+9$
21. ${c}^{2}-25$	23. ${b}^{2}-\dfrac{36}{49}$
25. $64{j}^{2}-16$	27. $81{c}^{2}-25$
29. $169-{q}^{2}$	31. $16-36{y}^{2}$
33. $49{w}^{2}-100{x}^{2}$	35. ${p}^{2}-\dfrac{16}{25}{q}^{2}$
37. ${x}^{2}{y}^{2}-81$	39. ${r}^{2}{s}^{2}-\dfrac{4}{49}$
41. $36{m}^{6}-16{n}^{10}$	43. $225{m}^{4}-64{n}^{8}$
45. a) $4{r}^{2}+48r+144$ b) $9{p}^{2}-64$ c) $7{a}^{2}-48ab-7{b}^{2}$ d) ${k}^{2}-12k+36$	47. a) ${x}^{10}-{y}^{10}$ b) ${m}^{6}-16{m}^{3}n+64{n}^{2}$ c) $81{p}^{2}+144pq+64{q}^{2}$ d) ${r}^{5}+{r}^{2}{s}^{2}-{r}^{3}{s}^{3}-{s}^{5}$
49. a) 4,225 b) 4,225 c) Answers will vary.	51. Answers will vary.
53. Answers will vary.

Attributions

This chapter has been adapted from “Special Products” in Elementary Algebra (OpenStax) by Lynn Marecek and MaryAnne Anthony-Smith, which is under a CC BY 4.0 Licence. Adapted by Izabela Mazur. See the Copyright page for more information.

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	$\left(5x-6\right)\left(6x+5\right)$
Use FOIL.	$30{x}^{2}+25x-36x-30$
Simplify.	$30{x}^{2}-11x-30$

Square a Binomial Using the Binomial Squares Pattern

Multiply Conjugates Using the Product of Conjugates Pattern

Recognize and Use the Appropriate Special Product Pattern

Key Concepts

Glossary

Practice Makes Perfect

Square a Binomial Using the Binomial Squares Pattern

Multiply Conjugates Using the Product of Conjugates Pattern

Recognize and Use the Appropriate Special Product Pattern

Everyday Math

Writing Exercises

Answers

Attributions

License

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