### 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

Before you get started, take this readiness quiz.

- Simplify: ⓐ ${9}^{2}$ ⓑ ${\left(\mathrm{-9}\right)}^{2}$ ⓒ $\text{\u2212}{9}^{2}.$

If you missed this problem, review Example 1.50.

### 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: | ${y}^{2}-14y+49$ |

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.

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!

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.

*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.

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

Putting it all together:

### Binomial Squares Pattern

If $a\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}b$ are real numbers,

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. | ${10}^{2}+\text{\_\_\_}+$ |

Square the last term. | ${10}^{2}+\text{\_\_\_}+{4}^{2}$ |

Double their product. | ${10}^{2}+2\xb710\xb74+{4}^{2}$ |

Simplify. | $100+80+16$ |

Simplify. | $196$ |

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

The pattern works!

### Example 6.47

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

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

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

### Example 6.48

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

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

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

### Example 6.49

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

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

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

### Example 6.50

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

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

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

### Example 6.51

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

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

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

### 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?

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

*Notice the first terms are the same in each pair.*

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

*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

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.

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.

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

This leads to the pattern:

### Product of Conjugates Pattern

If $a\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}b$ 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 conjudgates, so the result will be the difference of two squares. | $\text{\_\_\_\_}-\text{\_\_\_\_}$ |

Square the first term. | ${10}^{2}-\text{\_\_\_\_}$ |

Square the last term. | ${10}^{2}-{2}^{2}$ |

Simplify. | $100-4$ |

Simplify. | $96$ |

What do you get using the order of operations? | |

$\begin{array}{c}(10-2)(10+2)\\ (8)(12)\\ 96\end{array}$ |

Notice, the result is the same!

### Example 6.52

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

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

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

### Example 6.53

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

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

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

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 6.54

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

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

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

Now we’ll multiply conjugates that have two variables.

### Example 6.55

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

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

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

### Example 6.56

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

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

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

### Example 6.57

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

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

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

### 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 6.58

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

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

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

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

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

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

### Media

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

### Section 6.4 Exercises

#### Practice Makes Perfect

**Square a Binomial Using the Binomial Squares Pattern**

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

${(w+4)}^{2}$

${\left(y+\frac{1}{4}\right)}^{2}$

${(b-7)}^{2}$

${\left(m-15\right)}^{2}$

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

${(2q+\frac{1}{3})}^{2}$

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

${\left(\frac{1}{5}x-\frac{1}{7}y\right)}^{2}$

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

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

**Multiply Conjugates Using the Product of Conjugates Pattern**

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

$(m-7)(m+7)$

$\left(x+\frac{3}{4}\right)\left(x-\frac{3}{4}\right)$

$(5k+6)(5k-6)$

$(11k+4)(11k-4)$

$(11-b)(11+b)$

$(5-3x)(5+3x)$

$(9c-2d)(9c+2d)$

$\left(m+\frac{2}{3}n\right)\left(m-\frac{2}{3}n\right)$

$(ab-4)(ab+4)$

$(uv-\frac{3}{5})(uv+\frac{3}{5})$

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

$(12{p}^{3}-11{q}^{2})(12{p}^{3}+11{q}^{2})$

**Recognize and Use the Appropriate Special Product Pattern**

In the following exercises, find each product.

ⓐ $\left(p-3\right)\left(p+3\right)$

ⓑ ${\left(t-9\right)}^{2}$

ⓒ ${\left(m+n\right)}^{2}$

ⓓ $(2x+y)(x-2y)$

- ⓐ ${({a}^{5}-7b)}^{2}$
- ⓑ $({x}^{2}+8y)(8x-{y}^{2})$
- ⓒ $({r}^{6}+{s}^{6})({r}^{6}-{s}^{6})$
- ⓓ ${({y}^{4}+2z)}^{2}$

- ⓐ $({x}^{5}+{y}^{5})({x}^{5}-{y}^{5})$
- ⓑ ${({m}^{3}-8n)}^{2}$
- ⓒ ${(9p+8q)}^{2}$
- ⓓ $({r}^{2}-{s}^{3})({r}^{3}+{s}^{2})$

#### Everyday Math

**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 $(50-3)(50+3)$ by using the product of conjugates pattern, $\left(a-b\right)\left(a+b\right)={a}^{2}-{b}^{2}$.
- ⓑ Multiply $47\xb753$ without using a calculator.
- ⓒ Which way is easier for you? Why?

**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 ${(60+5)}^{2}$ 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?

#### Writing Exercises

How do you decide which pattern to use?

Marta did the following work on her homework paper:

Explain what is wrong with Marta’s work.

Use the order of operations to show that ${\left(3+5\right)}^{2}$ is 64, and then use that numerical example to explain why ${\left(a+b\right)}^{2}\ne {a}^{2}+{b}^{2}$.

#### 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?