## Multiplying Conjugates Using the Product of Conjugates Pattern

Contents

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?

\(\begin{array}{ccccccc}\hfill \left(x-9\right)\left(x+9\right)\hfill & & & \hfill \phantom{\rule{4em}{0ex}}\left(y-8\right)\left(y+8\right)\hfill & & & \hfill \phantom{\rule{4em}{0ex}}\left(2x-5\right)\left(2x+5\right)\hfill \end{array}\)

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

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

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.

\(\begin{array}{ccccccc}\hfill \left(x-9\right)\left(x+9\right)\hfill & & & \hfill \phantom{\rule{4em}{0ex}}\left(y-8\right)\left(y+8\right)\hfill & & & \hfill \phantom{\rule{4em}{0ex}}\left(2x-5\right)\left(2x+5\right)\hfill \\ \hfill {x}^{2}+9x-9x-81\hfill & & & \hfill \phantom{\rule{4em}{0ex}}{y}^{2}+8y-8y-64\hfill & & & \hfill \phantom{\rule{4em}{0ex}}4{x}^{2}+10x-10x-25\hfill \\ \hfill {x}^{2}-81\hfill & & & \hfill \phantom{\rule{4em}{0ex}}{y}^{2}-64\hfill & & & \hfill \phantom{\rule{4em}{0ex}}4{x}^{2}-25\hfill \end{array}\)

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.

\(\begin{array}{}\\ \hfill \left(a+b\right)\left(a-b\right)={a}^{2}-\text{____}\hfill \\ \hfill \text{To get the}\phantom{\rule{0.2em}{0ex}}\mathbf{\text{first term, square the first term}}.\hfill \end{array}\)

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

\(\begin{array}{}\\ \hfill \left(a+b\right)\left(a-b\right)={a}^{2}-{b}^{2}\hfill \\ \hfill \text{To get the}\phantom{\rule{0.2em}{0ex}}\mathbf{\text{last term, square the last term}}.\hfill \end{array}\)

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 \(a\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}b\) 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.

\(\begin{array}{cccc}& & & \phantom{\rule{4em}{0ex}}\left(10-2\right)\left(10+2\right)\hfill \\ \text{It is the product of conjugates, so the result will be the}\hfill & & & \\ \text{difference of two squares.}\hfill & & & \phantom{\rule{4em}{0ex}}\text{____}-\text{____}\hfill \\ \text{Square the first term.}\hfill & & & \phantom{\rule{4em}{0ex}}{10}^{2}-\text{____}\hfill \\ \text{Square the last term.}\hfill & & & \phantom{\rule{4em}{0ex}}{10}^{2}-{2}^{2}\hfill \\ \text{Simplify.}\hfill & & & \phantom{\rule{4em}{0ex}}100-4\hfill \\ \text{Simplify.}\hfill & & & \phantom{\rule{4em}{0ex}}96\hfill \\ \text{What do you get using the Order of Operations?}\hfill & & & \\ & & & \phantom{\rule{4em}{0ex}}\left(10-2\right)\left(10+2\right)\hfill \\ & & & \phantom{\rule{4em}{0ex}}\left(8\right)\left(12\right)\hfill \\ & & & \phantom{\rule{4em}{0ex}}96\hfill \end{array}\)

Notice, the result is the same!

## Example

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. | |

Square the first term, x. | |

Square the last term, 8. | |

The product is a difference of squares. |

## Example

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

### Solution

Are the binomials conjugates?

It is the product of conjugates. | |

Square the first term, 2x. | |

Square the last term, 5. | |

Simplify. The product is a difference of squares. |

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

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

### Solution

It is the product of conjugates. | |

Use the pattern. | |

Simplify. |

Now we’ll multiply conjugates that have two variables.

## Example

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

### Solution

This fits the pattern. | |

Use the pattern. | |

Simplify. |

## Example

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

### Solution

This fits the pattern. | |

Use the pattern. | |

Simplify. |

## Example

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

### Solution

This fits the pattern. | |

Use the pattern. | |

Simplify. |