Lesson Objectives

- Demonstrate an understanding of how to multiply polynomials
- Learn how to quickly multiply the sum and difference of two terms
- Learn how to quickly find the square of a binomial
- Learn how to quickly find the cube of a binomial

## Special Polynomial Products

In our last lesson, we learned how to multiply polynomials. In this lesson, we will show various formulas that can be used to find certain binomial products that occur frequently. We refer to these commonly occurring multiplication problems as "special polynomial products" or "special products".

(x + y)

(x - y)

We can see that the terms are the same, we have an x in each first position, and a y in each last position. The only difference between the two binomials is the sign. When this occurs, we can say the two binomials are "conjugates". When we multiply conjugates together, we can use a special formula:

(x + y)(x - y) = x

This means we can find the product of the first terms and subtract away the product of the last terms. Let's look at a few examples.

Example 1: Find each product.

(x - 6)(x + 6)

We can see that we are multiplying conjugates (sum and difference of the same two terms). Let's use our formula. This tells us to multiply the first terms and subtract away the product of the last terms:

(x - 6)(x + 6) = x

If we use the FOIL method, we get the same result. We will see when using FOIL that the middle terms drop out. This is due to the opposite signs that are involved.

(x - 6)(x + 6) = x

We can see that we obtain the same answer either way. Using the formula for multiplying conjugates is generally faster than using FOIL.

Example 2: Find each product.

(9x

We can see that we are multiplying conjugates (sum and difference of the same two terms). Let's use our formula. This tells us to multiply the first terms and subtract away the product of the last terms:

(9x

(x + y)

(x - y)

We give two different formulas to adjust for the sign. If we have a sum, we use the top formula, if we have a difference, we use the bottom formula. Let's look at a few examples.

Example 3: Find each product.

(x + 4)

Square the first term:

x

Find 2 times the first term times the second term:

2 • x • 4 = 8x

Square the last term:

4

Since our binomial is a sum, our answer is the sum of these three parts:

x

(x + 4)

(x + 4)

Example 4: Find each product.

(3x - 5y)

Square the first term:

(3x)

Find 2 times the first term times the second term:

2 • 3x • 5y = 30xy

Square the last term:

(5y)

Since our binomial is a difference, we will write our answer with a minus sign in front of the middle term:

9x

(3x - 5y)

(3x - 5y)

(x + y)

(x - y)

Let's look at a few examples.

Example 5: Find each product.

(2x + 5)

Cube the first term:

(2x)

Multiply 3 times the first term squared times the second term:

3 • (2x)

Multiply 3 times the first term times the second term squared:

3 • 2x • 5

Cube the last term:

5

Since our binomial is a sum, our answer is the sum of these four parts:

8x

(2x + 5)

(2x + 5)

Example 6: Find each product.

(7x - 4y)

Cube the first term:

(7x)

Multiply 3 times the first term squared times the second term:

3 • (7x)

Multiply 3 times the first term times the second term squared:

3 • 7x • (4y)

Cube the last term:

(4y)

Since our binomial is a difference, we will write our answer with a minus sign in front of the second and fourth terms:

343x

(7x - 4y)

(7x - 4y)

### Multiplying the Sum and Difference of Two Terms

When we have two binomials such as:(x + y)

(x - y)

We can see that the terms are the same, we have an x in each first position, and a y in each last position. The only difference between the two binomials is the sign. When this occurs, we can say the two binomials are "conjugates". When we multiply conjugates together, we can use a special formula:

(x + y)(x - y) = x

^{2}- y^{2}This means we can find the product of the first terms and subtract away the product of the last terms. Let's look at a few examples.

Example 1: Find each product.

(x - 6)(x + 6)

We can see that we are multiplying conjugates (sum and difference of the same two terms). Let's use our formula. This tells us to multiply the first terms and subtract away the product of the last terms:

(x - 6)(x + 6) = x

^{2}- 36If we use the FOIL method, we get the same result. We will see when using FOIL that the middle terms drop out. This is due to the opposite signs that are involved.

(x - 6)(x + 6) = x

^{2}+ 6x - 6x - 36 = x^{2}- 36We can see that we obtain the same answer either way. Using the formula for multiplying conjugates is generally faster than using FOIL.

Example 2: Find each product.

(9x

^{2}- 15y)(9x^{2}+ 15y)We can see that we are multiplying conjugates (sum and difference of the same two terms). Let's use our formula. This tells us to multiply the first terms and subtract away the product of the last terms:

(9x

^{2}- 15y)(9x^{2}+ 15y) = 81x^{4}- 225y^{2}### The Square of a Binomial

We will often need to square a binomial. When this situation occurs, we can use the following formula:(x + y)

^{2}= x^{2}+ 2xy + y^{2}(x - y)

^{2}= x^{2}- 2xy + y^{2}We give two different formulas to adjust for the sign. If we have a sum, we use the top formula, if we have a difference, we use the bottom formula. Let's look at a few examples.

Example 3: Find each product.

(x + 4)

^{2}Square the first term:

x

^{2}Find 2 times the first term times the second term:

2 • x • 4 = 8x

Square the last term:

4

^{2}= 16Since our binomial is a sum, our answer is the sum of these three parts:

x

^{2}+ 8x + 16(x + 4)

^{2}= x^{2}+ 2 • 4 • x + 4^{2}(x + 4)

^{2}= x^{2}+ 8x + 16Example 4: Find each product.

(3x - 5y)

^{2}Square the first term:

(3x)

^{2}= 9x^{2}Find 2 times the first term times the second term:

2 • 3x • 5y = 30xy

Square the last term:

(5y)

^{2}= 25y^{2}Since our binomial is a difference, we will write our answer with a minus sign in front of the middle term:

9x

^{2}- 30xy + 25y^{2}(3x - 5y)

^{2}= (3x)^{2}- 2 • 3x • 5y + (5y)^{2}(3x - 5y)

^{2}= 9x^{2}- 30xy + 25y^{2}### The Cube of a Binomial

Lastly, we want to cover the formula for finding the cube of a binomial. Just like with the formula for squaring a binomial, we have two different scenarios that change based on the sign.(x + y)

^{3}= x^{3}+ 3x^{2}y + 3xy^{2}+ y^{3}(x - y)

^{3}= x^{3}- 3x^{2}y + 3xy^{2}- y^{3}Let's look at a few examples.

Example 5: Find each product.

(2x + 5)

^{3}Cube the first term:

(2x)

^{3}= 8x^{3}Multiply 3 times the first term squared times the second term:

3 • (2x)

^{2}• 5 = 60x^{2}Multiply 3 times the first term times the second term squared:

3 • 2x • 5

^{2}= 150xCube the last term:

5

^{3}= 125Since our binomial is a sum, our answer is the sum of these four parts:

8x

^{3}+ 60x^{2}+ 150x + 125(2x + 5)

^{3}= (2x)^{3}+ 3 • (2x)^{2}• 5 + 3 • 2x • 5^{2}+ 5^{3}(2x + 5)

^{3}= 8x^{3}+ 60x^{2}+ 150x + 125Example 6: Find each product.

(7x - 4y)

^{3}Cube the first term:

(7x)

^{3}= 343x^{3}Multiply 3 times the first term squared times the second term:

3 • (7x)

^{2}• 4y = 588x^{2}yMultiply 3 times the first term times the second term squared:

3 • 7x • (4y)

^{2}= 336xy^{2}Cube the last term:

(4y)

^{3}= 64y^{3}Since our binomial is a difference, we will write our answer with a minus sign in front of the second and fourth terms:

343x

^{3}- 588x^{2}y + 336xy^{2}- 64y^{3}(7x - 4y)

^{3}= (7x)^{3}- 3 • (7x)^{2}• 4y + 3 • 7x • (4y)^{2}- (4y)^{3}(7x - 4y)

^{3}= 343x^{3}- 588x^{2}y + 336xy^{2}- 64y^{3}#### Skills Check:

Example #1

Simplify each.

Please choose the best answer. $$(3x - 8y)^{2}$$

A

$$9x^{2}- 48xy + 64y^{2}$$

B

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

C

$$9x^{2}+ 64y^{2}$$

D

$$9x^{2}- 64y^{2}$$

E

$$3x^{2}+ 16x - 24$$

Example #2

Simplify each. $$(y + 7x)(y - 7x)$$

Please choose the best answer.

A

$$y^{2}- 14xy + 49x^{2}$$

B

$$y - 16x$$

C

$$y^{2}+ 14xy + 49x^{2}$$

D

$$16y^{2}- 25x^{2}$$

E

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

Example #3

Simplify each. $$(2x + 1)^{3}$$

Please choose the best answer.

A

$$4x^{3}+ 2x^{2}+ x + 3$$

B

$$6x^{3}- 2x^{2}+ x - 1$$

C

$$24x^{2}- 11$$

D

$$8x^{3}+ 12x^{2}+ 6x + 1$$

E

$$8x^{3}- 12x^{2}+ 6x - 1$$

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