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Multiplying Conjugates Using the Product of Conjugates Pattern

Multiplying 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?

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

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

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.

This figure has three products. The first is x minus 9, in parentheses, times x plus 9, in parentheses. Below the x minus 9 is the word “difference”. Below x plus 9 is the word “sum”. The second is y minus 8, in parentheses, times y plus 8, in parentheses. Below y minus 8 is the word “difference”. Below y plus 8 is the word “sum”. The last is 2x minus 5, in parentheses, times 2x plus 5, in parentheses. Below the 2x minus 5 is the word “difference” and below 2x plus 5 is the word “sum”.

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.

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

This figure has three columns. The first column contains the product of x plus 9 and x minus 9. Below this is the expression x squared minus 9x plus 9x minus 81. Below this is x squared minus 81. The second column contains the product of y minus 8 and y plus 8. Below this is the expression y squared plus 8y minus 8y minus 64. Below this is y squared minus 64. The third column contains the product of 2x minus 5 and 2x plus 5. Below this is the expression 4x squared plus 10x minus 10x minus 25. Below this is 4x squared minus 25.

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,

This figure is divided into two sides. On the left side is the following formula: the product of a minus b and a plus b equals a squared minus b squared. On the right side is the same formula labeled: a minus b and a plus b are labeled “conjugates”, the a squared and b squared are labeled squares and the minus sign between the squares is labeled “difference”. Therefore, the product of two conjugates is called a difference of squares.

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. The product of x minus 8 and x plus 8. Above this is the general form a minus b, in parentheses, times a plus b, in parentheses.
Square the first term, x. x squared minus blank. Above this is the general form a squared minus b squared.
Square the last term, 8. x squared minus 8 squared.
The product is a difference of squares. x squared minus 64.

Example

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

Solution

Are the binomials conjugates?

It is the product of conjugates. The product of 2x plus 5 and 2x minus 5. Above this is the general form a minus b, in parentheses, times a plus b, in parentheses.
Square the first term, 2x. 2 x squared minus blank. Above this is the general form a squared minus b squared.
Square the last term, 5. 2 x squared minus 5 squared.
Simplify. The product is a difference of squares. 4 x squared minus 25.

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. The product of 3 plus 5 x and 3 minus 5 x. Above this is the general form a plus b, in parentheses, times a minus b, in parentheses.
Use the pattern. 3 squared minus 5 x squared. Above this is the general form a squared minus b squared.
Simplify. 9 minus 25 x squared.

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. 5 m minus 9 n and 5 m plus 9 n. Above this is the general form a plus b, in parentheses, times a minus b, in parentheses.
Use the pattern. 5 m squared minus 9 n squared. Above this is the general form a squared minus b squared.
Simplify. 25 m squared minus 81 n squared.

Example

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

Solution

This fits the pattern. The product of c d minus 8 and c d plus 8. Above this is the general form a plus b, in parentheses, times a minus b, in parentheses.
Use the pattern. c d squared minus 8 squared. Above this is the general form a squared minus b squared.
Simplify. c squared d squared minus 64.

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. The product of 6 u squared minus 11 v to the fifth power and 6 u squared plus 11 v to the fifth power. Above this is the general form a plus b, in parentheses, times a minus b, in parentheses.
Use the pattern. 6 u squared, in parentheses, squared, minus 11 v to the fifth power, in parentheses, squared. Above this is the general form a squared minus b squared.
Simplify. 36 u to the fourth power minus 121 v to the tenth power.

This lesson is part of:

Polynomials II

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