Lesson Objectives

- Demonstrate an understanding of multiplication with polynomials
- Learn how to quickly find the product of a binomial squared
- Learn how to quickly find the product of the sum and difference of two terms
- Learn how to quickly find the product of a binomial cubed

## What are Special Products of Polynomials

When working with polynomial multiplication, we have special products' formulas that can be applied to various problems involving binomials.

(x - y)

When we square a binomial, we get a trinomial that contains:

The square of the first term

Twice the product of the two terms

The square of the final term

The sign of the first term and the last term will always be positive. This is the result of squaring a positive or negative value. The sign of the middle term depends on the operation ("+" or "-") inside of the binomial. Let's look at the two cases:

(x + y)

(x - y)

We can see that squaring a sum (x + y) leads to a trinomial where all terms are positive. When we square a difference (x - y), the middle term is negative. Let's look at some examples.

Example 1: Find each product.

(x + 5)

1) Square the first term:

x

2) Find the product of 2 times the two terms:

2 • x • 5 = 10x

3) Square the final term:

5

In this case, we have a binomial with "+", therefore, we use a "+" sign for the middle term in our trinomial:

x

(x + 5)

Example 2: Find each product.

(3x - 2)

1) Square the first term:

(3x)

2) Find the product of 2 times the two terms:

2 • 3x • 2 = 12x

3) Square the final term:

2

In this case, we have a binomial with "-", therefore, we use a "-" sign for the middle term in our trinomial:

9x

(x + y) and (x - y)

We can see that only the sign differs. The first term in each case is (x). The second term in each case is (y). When this scenario occurs, we can say we have conjugates. We can also say we have the product of the sum and difference of two terms. When we multiply conjugates, we have a special products' formula that allows us to shorten the process:

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

To think about why this formula works, let's use FOIL to find our product:

(x + y)(x - y)

F » x • x = x

O » x • -y = -xy

I » x • y = xy

L » y • -y = -y

We can see that our two middle terms will cancel out. This will always be the case when multiplying conjugates. We are just left with the product of the first terms (x

Example 3: Find each product.

(x + 7)(x - 7)

1) Square the first term:

x

2) Square the second term:

7

3) The result is the square of the first term (x

x

(x + 7)(x - 7) = x

Example 4: Find each product.

(11x + 4y)(11x - 4y)

1) Square the first term:

(11x)

2) Square the second term:

(4y)

3) The result is the square of the first term (121x

121x

(11x + 4y)(11x - 4y) = 121x

(x - y)

When we cube a binomial, we get a four-term polynomial that contains:

The first term cubed

Three multiplied by the first term squared and the second term

Three multiplied by the first term and the second term squared

The last term cubed

The sign of the first term and third term of the answer will always be positive. The sign of the second term and fourth term will change based on the sign in the binomial being cubed. Let's look at the two cases:

(x + y)

(x - y)

When we cube a sum (x + y), we obtain a four-term polynomial where all terms are positive. When we cube a difference (x - y), we obtain a four-term polynomial where the second and fourth terms are negative. Let's look at some examples.

Example 5: Find each product.

(x + 4)

1) Cube the first term:

x

2) Find the product of 3 times the first term squared and the second term:

3 • x

3) Find the product of 3 times the first term and the second term squared:

3 • x • 4

4) Cube the final term:

4

In this case, we have a binomial with "+", therefore, we use a "+" sign for all terms:

x

(x + 4)

Example 6: Find each product.

(2x - 5)

1) Cube the first term:

(2x)

2) Find the product of 3 times the first term squared and the second term:

3 • (2x)

3) Find the product of 3 times the first term and the second term squared:

3 • 2x • 5

4) Cube the final term:

5

In this case, we have a binomial with "-", therefore, we use a "-" sign for the second and fourth terms:

8x

(2x - 5)

### The Square of a Binomial

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

^{2}= x^{2}- 2xy + y^{2}When we square a binomial, we get a trinomial that contains:

The square of the first term

Twice the product of the two terms

The square of the final term

The sign of the first term and the last term will always be positive. This is the result of squaring a positive or negative value. The sign of the middle term depends on the operation ("+" or "-") inside of the binomial. Let's look at the two cases:

(x + y)

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

^{2}= x^{2}- 2xy + y^{2}We can see that squaring a sum (x + y) leads to a trinomial where all terms are positive. When we square a difference (x - y), the middle term is negative. Let's look at some examples.

Example 1: Find each product.

(x + 5)

^{2}1) Square the first term:

x

^{2}2) Find the product of 2 times the two terms:

2 • x • 5 = 10x

3) Square the final term:

5

^{2}= 25In this case, we have a binomial with "+", therefore, we use a "+" sign for the middle term in our trinomial:

x

^{2}+ 10x + 25(x + 5)

^{2}= x^{2}+ 10x + 25Example 2: Find each product.

(3x - 2)

^{2}1) Square the first term:

(3x)

^{2}= 3^{2}x^{2}= 9x^{2}2) Find the product of 2 times the two terms:

2 • 3x • 2 = 12x

3) Square the final term:

2

^{2}= 4In this case, we have a binomial with "-", therefore, we use a "-" sign for the middle term in our trinomial:

9x

^{2}- 12x + 4### Product of Conjugates - Product of the Sum and Difference of Two Terms

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

We can see that only the sign differs. The first term in each case is (x). The second term in each case is (y). When this scenario occurs, we can say we have conjugates. We can also say we have the product of the sum and difference of two terms. When we multiply conjugates, we have a special products' formula that allows us to shorten the process:

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

^{2}- y^{2}To think about why this formula works, let's use FOIL to find our product:

(x + y)(x - y)

F » x • x = x

^{2}O » x • -y = -xy

I » x • y = xy

L » y • -y = -y

^{2}We can see that our two middle terms will cancel out. This will always be the case when multiplying conjugates. We are just left with the product of the first terms (x

^{2}) and the product of the last terms (-y^{2}). Let's look at a few examples.Example 3: Find each product.

(x + 7)(x - 7)

1) Square the first term:

x

^{2}2) Square the second term:

7

^{2}= 493) The result is the square of the first term (x

^{2}) less (-) the square of the second term (49):x

^{2}- 49(x + 7)(x - 7) = x

^{2}- 49Example 4: Find each product.

(11x + 4y)(11x - 4y)

1) Square the first term:

(11x)

^{2}= 11^{2}x^{2}= 121x^{2}2) Square the second term:

(4y)

^{2}= 4^{2}y^{2}= 16y^{2}3) The result is the square of the first term (121x

^{2}) less (-) the square of the second term (16y^{2}):121x

^{2}- 16y^{2}(11x + 4y)(11x - 4y) = 121x

^{2}- 16y^{2}### Cube of a Binomial

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

^{3}= x^{3}- 3x^{2}y + 3xy^{2}- y^{3}When we cube a binomial, we get a four-term polynomial that contains:

The first term cubed

Three multiplied by the first term squared and the second term

Three multiplied by the first term and the second term squared

The last term cubed

The sign of the first term and third term of the answer will always be positive. The sign of the second term and fourth term will change based on the sign in the binomial being cubed. Let's look at the two cases:

(x + y)

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

^{3}= x^{3}- 3x^{2}y + 3xy^{2}- y^{3}When we cube a sum (x + y), we obtain a four-term polynomial where all terms are positive. When we cube a difference (x - y), we obtain a four-term polynomial where the second and fourth terms are negative. Let's look at some examples.

Example 5: Find each product.

(x + 4)

^{3}1) Cube the first term:

x

^{3}2) Find the product of 3 times the first term squared and the second term:

3 • x

^{2}• 4 = 12x^{2}3) Find the product of 3 times the first term and the second term squared:

3 • x • 4

^{2}= 3 • x • 16 = 48x4) Cube the final term:

4

^{3}= 64In this case, we have a binomial with "+", therefore, we use a "+" sign for all terms:

x

^{3}+ 12x^{2}+ 48x + 64(x + 4)

^{3}= x^{3}+ 12x^{2}+ 48x + 64Example 6: Find each product.

(2x - 5)

^{3}1) Cube the first term:

(2x)

^{3}= 2^{3}x^{3}= 8x^{3}2) Find the product of 3 times the first term squared and the second term:

3 • (2x)

^{2}• 5 = 3 • 2^{2}x^{2}• 5 = 3 • 4x^{2}• 5 = 60x^{2}3) Find the product of 3 times the first term and the second term squared:

3 • 2x • 5

^{2}= 3 • 2x • 25 = 150x4) Cube the final term:

5

^{3}= 125In this case, we have a binomial with "-", therefore, we use a "-" sign for the second and fourth terms:

8x

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

^{3}= 8x^{3}- 60x^{2}+ 150x - 125#### Skills Check:

Example #1

Find each product. $$(5x - 2)^{2}$$

Please choose the best answer.

A

$$25x^{4}+ 4$$

B

$$25x^{2}+ 80x + 64$$

C

$$25x^{2}- 4$$

D

$$10x^{2}- 25x + 4$$

E

$$25x^{2}- 20x + 4$$

Example #2

Find each product. $$(8x + 7)(8x - 7)$$

Please choose the best answer.

A

$$64x^{2}- 49$$

B

$$36x^{2}+ 60x + 25$$

C

$$64x^{2}+ 112x + 49$$

D

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

E

$$64x^{2}- 15x + 49$$

Example #3

Find each product. $$(x - 9)^{3}$$

Please choose the best answer.

A

$$x^{3}- 729$$

B

$$x^{3}- 27x^{2}+ 243x - 729$$

C

$$x^{3}- 81x^{2}- 729$$

D

$$3x^{3}+ 81x^{2}- 243x + 81$$

E

$$3x^{3}- 729$$

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