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# 9.2: Multiplication of Polynomials

Created by: CK-12

## Learning Objectives

• Multiply a polynomial by a monomial
• Multiply a polynomial by a binomial
• Solve problems using multiplication of polynomials

## Introduction

When multiplying polynomials we must remember the exponent rules that we learned in the last chapter.

The Product Rule $x^n \cdot x^m = x^{n + m}$

This says that if we multiply expressions that have the same base, we just add the exponents and keep the base unchanged.

If the expressions we are multiplying have coefficients and more than one variable, we multiply the coefficients just as we would any number and we apply the product rule on each variable separately.

$(2x^2y^3) \cdot (3x^2y) = (2 \cdot 3) \cdot (x^{2 + 2}) \cdot (y^{3 + 1}) = 6x^4y^4$

## Multiplying a Polynomial by a Monomial

We begin this section by multiplying a monomial by a monomial. As you saw above, we need to multiply the coefficients separately and then apply the exponent rules to each variable separately. Let’s try some examples.

Example 1

Multiply the following monomials.

a) $(2x^2) (5x^3)$

b) $(-3y^4) (2y^2)$

c) $(3xy^5) (-6x^4y^2)$

d) $(-12a^2b^3c^4) (-3a^2b^2)$

Solution

a) $(2x^2) (5x^3) = (2 \cdot 5) \cdot (x^2 \cdot x^3) = 10x^{2 + 3} = 10x^5$

b) $(-3y^4) (2y^2) = (-3 \cdot 2) \cdot (y^4 \cdot y^2) = -6y^{4 + 2} = -6y^6$

c) $(3xy^5) (-6x^4y^2) = 18x^{1 + 4}y^{5 + 2} = -18x^5y^7$

d) $(-12a^2b^3c^4)(-3a^2b^2) = 36a^{2 + 2} b^{3 + 2} c^4 = 36a^4b^5c^4$

To multiply a polynomial by a monomial, we use the Distributive Property.

This says that

$a(b + c) = a b + a c$

This property is best illustrated by an area problem. We can find the area of the big rectangle in two ways.

One way is to use the formula for the area of a rectangle.

$\text{Area of the big rectangle} & = \text{length} \cdot \ \text{width }\\\text{Length} & = a, \ \text{Width} = b + c \\\text{Area} & = a \cdot (b \cdot c)$

The area of the big rectangle can also be found by adding the areas of the two smaller rectangles.

$\text{Area of red rectangle} & = ab \\\text{Area of blue rectangle} & = ac \\\text{Area of big rectangle} & = ab + ac$

This means that $a(b + c) = ab + ac$. It shows why the Distributive Property works.

This property is useful for working with numbers and also with variables.

For instance, to solve this problem, you would add 2 and 7 to get 9 and then multiply by 5 to get 45. But there is another way to do this.

$5(2 + 7) = 5 \cdot 2 + 5 \cdot 7$

It means that each number in the parenthesis is multiplied by 5 separately and then the products are added together.

$5(2 + 7) = 5 \cdot 2 + 5 \cdot 7 = 10 + 35 = 45$

In general, if we have a number or variable in front of a parenthesis, this means that each term in the parenthesis is multiplied by the expression in front of the parenthesis. The distributive property works no matter how many terms there are inside the parenthesis.

$a(b + c + d + e + f + \ldots) = a . b + a . c + a . d + a . e + a . f + \ldots$

The “...” means “and so on”.

Let’s now apply this property to multiplying polynomials by monomials.

Example 2

Multiply

a) $3(x^2 + 3x - 5)$

b) $4x(3x^2 - 7)$

c) $-7y(4y^2 - 2y + 1)$

Solution

a) $3(x^2 + 3x - 5) = 3(x^2) + 3(3x) - 3(5) = 3x^2 + 9x -15$

b) $4x(3x^2 - 7) = (4x)(3x^2) + (4x)(-7) = 12x^3 - 28x$

c) $-7y(4y^2 - 2y + 1) = (-7y) (4y^2) + (-7y) (-2y) + (-7y) (1) = -28y^3 + 14y^2 - 7y$

Notice that the use of the Distributive Property simplifies the problems to just multiplying monomials by monomials and adding all the separate parts together.

Example 3

Multiply

a) $2x^3(-3x^4 + 2x^3 -10x^2 + 7x +9)$

b) $-7a^2 bc ^3(5a^2 - 3b^2 - 9c^2)$

Solution

a) $2x^3(-3x^4 + 2x^3 -10x^2 + 7x +9) & = (2x^3)(-3x^4) + (2x^3)(2x^3) + (2x^3)(-10x^2) + (2x^3) (7x) + (2x^3) (9) \\& =-6x^7 + 4x^6 -20x^5 + 14x^4 + 18x^3$

b) $-7a^2 bc^3(5a^2 - 3b^2 - 9c^2) & =(-7a^2 bc^3)(5a^2) + (-7a^2 bc^3)(-3b^2) + (-7a^2 bc^3)(-9c^2) \\& = -35a^4 bc^3 + 21a^2b^3c^3 + 63a^2 bc^5$

## Multiply a Polynomial by a Binomial

Let’s start by multiplying two binomials together. A binomial is a polynomial with two terms, so a product of two binomials will take the form.

$(a + b) (c + d)$

The Distributive Property also applies in this situation. Let’s think of the first parenthesis as one term. The Distributive Property says that the term in front of the parenthesis multiplies with each term inside the parenthesis separately. Then, we add the results of the products.

$(a + b) (c + d) = (a + b) \cdot c + (a + b) \cdot d$

Let’s rewrite this answer as $c \cdot (a + b) + d \cdot (a + b)$

We see that we can apply the distributive property on each of the parenthesis in turn.

$c \cdot (a + b) + d \cdot (a + b) = c \cdot a + c \cdot b + d \cdot a + d \cdot b\ (\text{or}\ c a + c b + d a + d b)$

What you should notice is that when multiplying any two polynomials, every term in one polynomial is multiplied by every term in the other polynomial.

Let’s look at some examples of multiplying polynomials.

Example 4

Multiply and simplify $(2x + 1) (x + 3)$

Solution

We must multiply each term in the first polynomial with each term in the second polynomial.

Let’s try to be systematic to make sure that we get all the products.

First, multiply the first term in the first parenthesis by all the terms in the second parenthesis.

We are now done with the first term.

Now we multiply the second term in the first parenthesis by all terms in the second parenthesis and add them to the previous terms.

We are done with the multiplication and we can simplify.

$(2x) (x) + (2x) (3) + (1) (x) + (1) (3) & = 2x^2 + 6x + x + 3 \\& = 2x^2 + 7x + 3$

This way of multiplying polynomials is called in-line multiplication or horizontal multiplication.

Another method for multiplying polynomials is to use vertical multiplication similar to the vertical multiplication you learned with regular numbers. Let’s demonstrate this method with the same example.

$& \quad \quad \quad 2x \ + \ 1\\& \underline{\;\;\;\;\;\;\;\;\;\;\;\;x \ \ + \ 3}\\& \quad \quad \quad 6x \ + \ 3 \ \leftarrow \ \text{Multiply each term on top by}\ 3\\\text{Multiply each term on top by}\ x \rightarrow & \underline{2x^2 \ + \ x\;\;\;\;\;\;\;\;\;\;}\\& 2x^2 \ + \ 7x \ + \ 3 \leftarrow \ \text{Arrange like terms on top of each other}\\& \qquad \qquad \qquad \qquad \quad \text{and add vertically}$

This method is typically easier to use although it does take more space. Just make sure that like terms are together in vertical columns so you can combine them at the end.

Example 5

Multiply and simplify

(a) $(4x - 5) (x - 20)$

(b) $(3x - 2) (3x + 2)$

(c) $(3x^2 + 2x - 5) (2x - 3)$

(d) $(x^2 - 9) (4x^4 + 5x^2 - 2)$

Solution

a) $(4x - 5) (x - 20)$

Horizontal multiplication

$(4x - 5) (x - 20) & = (4x)(x) + (4x)(-20) + (-5)(x) + (-5)(-20) \\& = 4x^2 - 80x - 5x + 100 = 4x^2 - 85x + 100$

Vertical multiplication

Arrange the polynomials on top of each other with like terms in the same columns.

$& \quad \quad \quad \ 4x \ \ - \ \ \ 5\\& \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;x \ \ - \ \ 20}\\& \quad \quad -80x \ + \ 100\\& \underline{4x^2 \ - \ 5x\;\;\;\;\;\;\;\;\;\;\;\;\;}\\& 4x^2 \ - \ 85x \ + \ 100$

Both techniques result in the same answer, $4x^2 - 85x + 100.$

For the last question, we’ll show the solution with vertical multiplication because it may be a technique you are not used to. Horizontal multiplication will result in the exact same terms and the same answer.

b) $(3x - 2) (3x + 2)$

$& \quad \quad \quad 3x \ \ - \ 2\\& \underline{\;\;\;\;\;\;\;\;\;\;3x \ \ + \ 2}\\& \quad \quad \quad 6x \ - \ 4\\& \underline{9x^2 \ - 6x\;\;\;\;\;\;\;\;\;\;}\\& 9x^2 \ + 0x \ - 4$

Answer $9x^2 - 4$

(c) $(3x^2 + 2x -5) (2x - 3)$

It’s better to place the smaller polynomial on the bottom:

$& \quad \quad \quad 3x^{2} + 2x - 5\\& \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ 2x - 3}\\& \quad \ -9x^2 - 6x + 15\\& \underline{6x^3 + 4x^2 - 10x\;\;\;\;\;\;\;}\\& 6x^3 - 5x^{2} - 16x + 15$

Answer $6x^3 - 5x^2 - 16x + 15$

(d) $(x^2 - 9) (4x^4 + 5x^2 - 2)$

Set up the multiplication vertically and leave gaps for missing powers of $x$:

$& \quad \quad \quad 4x^4 \ + \ 5x^2 \ - \ 2\\& \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;x^2 \ - \ 9}\\& \quad \ \ -36x^4 - \ 45x^2 + 18\\& \underline{4x^6 + \ 5x^4 \ - \ 2x^2 \;\;\;\;\;\;\;\;\;\;}\\& 4x^ 6 - 31x^4 - 47x^2 + 18$

Answer $4x^6 - 31x^4 - 47x^2 + 18$

Multimedia Link The following video shows how multiplying two binomials together is related to the distributive property. Khan Academy Multiplying Expressions (7:59)

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## Solve Real-World Problems Using Multiplication of Polynomials

In this section, we will see how multiplication of polynomials is applied to finding the areas and volumes of geometric shapes.

Example 6

Find the areas of the following figures

a)

b)

Find the volumes of the following figures

c)

d)

Solution

a) We use the formula for the area of a rectangle.

$\text{Area} = \text{length} \cdot \ \text{width}$

For the big rectangle

$\text{Length} & = b + 3, \ \text{Width} = b + 2\\\text{Area} & = (b + 3)(b + 2)\\& = b^2 + 2b + 3b + 6\\& = b^2 + 5b + 6$

b) Let’s find the area of the big rectangle in the second figure and subtract the area of the yellow rectangle.

$\text{Area of big rectangle} & = 20(12) = 240\\\text{Area of yellow rectangle} & = (12- x)(20 - 2x)\\& = 240 - 24x - 20x + 2x^2\\& = 240 - 44x + 2x^2\\& = 2x^2 - 44x + 240$

The desired area is the difference between the two.

$\text{Area} & = 240 - (2x^2 - 44x + 240)\\& = 240 + (-2x^2 + 44x - 240)\\& = 240 - 2x^2 + 44x - 240\\& = -2x^2 + 44x$

c) The volume of this shape = (area of the base) $\cdot$ (height).

$\text{Area of the base } & = x(x + 2)\\& = x^2 + 2x\\\text{Height}& = 2x + 1\\\text{Volume} & = (x^2 + 2x) (2x + 1)\\& = 2x^3 + x^2 + 4x^2 + 2x\\& = 2x^3 + 5x^2 + 2x$

d) The volume of this shape = (area of the base) $\cdot$ (height).

$\text{Area of the base} & = (4a - 3) (2a + 1)\\& = 8a^2 + 4a - 6a - 3\\& = 8a^2 - 2a - 3\\\text{Height} & = a + 4 \\\text{Volume} & = (8a^2 - 2a - 3) (a + 4)$

Let’s multiply using the vertical method:

$& \quad \quad \quad \quad 8a^2 \ - 2a \ - \ 3\\& \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;a \ + \ 4}\\& \quad \quad \quad 32a^2 \ - \ 8a \ - \ 12\\& \underline{8a^3 \ - \ 2a^2 \ - \ 3a\;\;\;\;\;\;\;\;\;\;\;\;}\\& 8a^3 \ + \ 30a^2 - 11a \ - \ 12$

Answer $\text{Volume} = 8a^3 + 30a^2 - 11a -12$

## Review Questions

Multiply the following monomials.

1. $(2x) (-7x)$
2. $(-5a^2b) (-12a^3b^3)$
3. $(3xy^2 z^2) (15x^2 yz^3)$

Multiply and simplify.

1. $2x(4x - 5)$
2. $9x^3 (3x^2 - 2x + 7)$
3. $-3a^2b(9a^2 - 4b^2)$
4. $(x - 3) (x + 2)$
5. $(a^2 + 2) (3a^2 - 4)$
6. $(7x - 2) (9x - 5)$
7. $(2x - 1) (2x^2 - x + 3)$
8. $(3x + 2) (9x^2 - 6x + 4)$
9. $(a^2 + 2a - 3) (a^2 - 3a + 4)$

Find the areas of the following figures.

Find the volumes of the following figures.

1. $- 14x^2$
2. $60a^5b^4$
3. $45x^3y^3 z^5$
4. $8x^2 - 10x$
5. $27x^5 - 18x^4 + 63x^3$
6. $-27a^4b + 12a^2b^3$
7. $x^2 - x - 6$
8. $3a^4 + 2a^2 - 8$
9. $63x^2 - 53x + 10$
10. $4x^3 - 4x^2 + 7x - 3$
11. $27x^3 + 8$
12. $a^4 - a^3 - 5a^2 + 17a - 12$
13. $(2x + 4)(x + 6) = 2x^2 +16x + 24$
14. $x(3x + 8) = 3x^2 + 8x$
15. $6x^3 + 14x^2 + 8x$
16. $24x^3 - 28x^2 - 12x$

Feb 22, 2012

Aug 04, 2014