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Polynomials are useful for modeling different real-life situations. Sometimes more than one polynomial could be involved and the polynomials may need to be multiplied. For such a reason, this lesson aims to teach different methods for multiplying polynomials.

Catch-Up and Review

Here are a few recommended readings before getting started with this lesson.

Polynomial
Standard Form of a Polynomial
Like Terms
Addition and Subtraction of Polynomials
Product of Powers Property
Distributive Property

Challenge

Investigating the Product of Polynomials

When two polynomials are added or subtracted, the result is also a polynomial. What about the product of two polynomials? Consider, for example, the following pair of polynomials.

P (x) Q (x) = 2 x^{3} - x + 5 = 4 (x + 2) (x - 1)

Is it possible to calculate the product of

P (x)

and

Q (x) ?

P (x) \cdot Q (x)

also a polynomial? In the affirmative case, what are the degree, the leading coefficient, and the constant term of the resulting polynomial?

Discussion

Multiplying a Polynomial by a Constant

Consider a polynomial

P (x)

written in standard form.

P (x) = a_{n} x^{n} + a_{n - 1} x^{n - 1} + \dots + a_{1} x + a_{0}

Using the Distributive Property, multiply

P (x)

by a non-zero constant

C .

P (x) = a_{n} x^{n} + a_{n - 1} x^{n - 1} + \dots + a_{1} x + a_{0}

MultEqn

$LHS \cdot C = RHS \cdot C$

C \cdot P (x) = C (a_{n} x^{n} + a_{n - 1} x^{n - 1} + \dots + a_{1} x + a_{0})

Distr

Distribute $C$

C \cdot P (x) = C \cdot a_{n} x^{n} + C \cdot a_{n - 1} x^{n - 1} + \dots + C \cdot a_{1} x + C \cdot a_{0}

Note that the right-hand side is a polynomial of the same degree as

P (x)

— the degree of both is

n .

On the other hand, if

C \neq = 1,

the leading coefficient of

C \cdot P (x)

is different from the leading coefficient of

P (x) .

Polynomial	Degree	Leading Coefficient
$P (x) = a_{n} x^{n} + a_{n - 1} x^{n - 1} + \dots + a_{1} x + a_{0}$	$n$	$a_{n}$
$C \cdot P (x) = C \cdot a_{n} x^{n} + C \cdot a_{n - 1} x^{n - 1} + \dots + C \cdot a_{1} x + C \cdot a_{0}$	$n$	$C \cdot a_{n}$

Consequently, when a polynomial

P (x)

is multiplied by a constant

C,

the degree of the polynomial does not change and the leading coefficient equals

C

times the leading coefficient of

P (x) .

Example

Finding the Area of a Poster

Ali is on his school's student council. He is in charge of decorations for an upcoming school dance. He had the following chat with Kevin.

A: Hi Kevin. I need a poster to hang on the wall by the gym. K: Hello Ali. How big is the wall? A: I do not remember the width, but I know it is 10 feet high. K: Would you like to cover the entire wall with the poster? A: No. I would like there to be 1 foot of margin on each side of the poster. K: Ok. Once you tell me the width of the wall, I can find the cost of the poster. Thanks!

After the talk, Kevin made the following diagram.

A rectangle with dimensions 10 ft by x ft. A smaller rectangle inside with dimensions 8 ft by (x-2) ft.

a Write a polynomial

A (x),

in standard form, that models the area of the poster.

b What are the degree and leading coefficient of

A (x) ?

c The next day, Ali called Kevin and told him that the wall is

16

feet wide. What area will the poster have?

Hint

a The poster is

2

feet shorter and narrower than the wall. To find its area, multiply the length by the width.

b The degree of a polynomial is the greatest exponent in the polynomial. The leading coefficient is the coefficient in front of the term with the greatest exponent.

c Evaluate

A (x)

x = 16 .

Solution

a The area of the poster equals its length multiplied by its width.

A = ℓ \cdot w

From the chat and the diagram made by Kevin, the poster has to be

2

feet shorter and narrower than the wall. Since the wall is

10

feet tall, the poster will be

8

feet tall. The wall is

x

feet wide, so the width of the poster is

x - 2

feet.

A polynomial modeling the area of the poster is obtained by substituting the poster's dimensions into the formula for the area.

A (x) = (x - 2) \cdot 8 ⇓ A (x) = 8 x - 16

Note that this polynomial gives the area of the poster depending on the width of the wall.

b Start by writing the polynomial found in Part A. Then, highlight the term with the greatest exponent and its coefficient.

A (x) = 8 x - 16 ⇕ A (x) = 8 x^{1} - 16

The greatest exponent is

1,

which means that the degree of the polynomial is

1 .

The coefficient in front of the term with the greatest exponent is

8 .

Thus, the leading coefficient is

8 .

c Since

x

represents the width of the wall, to find the area of the poster, substitute

x = 16

into

A (x) .

A (x) = 8 x - 16

Substitute

$x = 16$

A (16) = 8 (16) - 16

Multiply

A (16) = 128 - 16

SubTerms

Subtract terms

A (16) = 112

The area of the poster will be

112

square feet.

Discussion

Methods for Multiplying Polynomials

Different methods can be used to multiply two polynomials. The following three methods are based on the Distributive Property.

Method

Multiplying Polynomials Using the Distributive Property

The product of two polynomials can be calculated by using the Distributive Property. Consider, for example, the following pair of polynomials.

P (x) Q (x) = x^{3} + 2 x^{2} - 3 = x^{2} + 4

To multiply these two polynomials, the following four steps can be followed.

Distribute One Polynomial to All the Terms of the Other

Start by writing the product

P (x) \cdot Q (x) .

(x^{3} + 2 x^{2} - 3) (x^{2} + 4)

Next, distribute

P (x)

to each term of

Q (x) .

(x^{3} + 2 x^{2} - 3) (x^{2} + 4)

Distr

Distribute $x^{3} + 2 x^{2} - 3$

(x^{3} + 2 x^{2} - 3) x^{2} + (x^{3} + 2 x^{2} - 3) 4

Clear Parentheses by Applying the Distributive Property

Apply the Distributive Property again to clear all the parentheses.

(x^{3} + 2 x^{2} - 3) x^{2} + (x^{3} + 2 x^{2} - 3) 4

Distr

Distribute $x^{2}$

(x^{3} \cdot x^{2} + 2 x^{2} \cdot x^{2} - 3 \cdot x^{2}) + (x^{3} + 2 x^{2} - 3) 4

Distr

Distribute $4$

(x^{3} \cdot x^{2} + 2 x^{2} \cdot x^{2} - 3 \cdot x^{2}) + (x^{3} \cdot 4 + 2 x^{2} \cdot 4 - 3 \cdot 4)

Apply the Product of Powers Property

Use the Product of Powers Property to rewrite the product of two or more exponents as a single power.

(x^{3} \cdot x^{2} + 2 x^{2} \cdot x^{2} - 3 \cdot x^{2}) + (x^{3} \cdot 4 + 2 x^{2} \cdot 4 - 3 \cdot 4)

MultPow

$a^{m} \cdot a^{n} = a^{m + n}$

(x^{3 + 2} + 2 x^{2 + 2} - 3 \cdot x^{2}) + (x^{3} \cdot 4 + 2 x^{2} \cdot 4 - 3 \cdot 4)

AddTerms

Add terms

(x^{5} + 2 x^{4} - 3 \cdot x^{2}) + (x^{3} \cdot 4 + 2 x^{2} \cdot 4 - 3 \cdot 4)

Combine Like Terms and Simplify

Finally, combine like terms and perform all the required operations to simplify the result.

(x^{5} + 2 x^{4} - 3 \cdot x^{2}) + (x^{3} \cdot 4 + 2 x^{2} \cdot 4 - 3 \cdot 4)

Multiply

(x^{5} + 2 x^{4} - 3 x^{2}) + (4 x^{3} + 8 x^{2} - 12)

CommutativePropAdd

Commutative Property of Addition

x^{5} + 2 x^{4} + 4 x^{3} - 3 x^{2} + 8 x^{2} - 12

AssociativePropAdd

Associative Property of Addition

x^{5} + 2 x^{4} + 4 x^{3} + (- 3 x^{2} + 8 x^{2}) - 12

AddTerms

Add terms

x^{5} + 2 x^{4} + 4 x^{3} + 5 x^{2} - 12

Note that multiplying a polynomial with

n

terms by a polynomial with

m

terms produces

n \cdot m

products. When two polynomials are multiplied, the product is a new polynomial whose degree equals the sum of the degrees of the polynomial factors.

$Animation showing the product of two polynomials in standard form. The degree of the product is d_1+d_2. The leading coefficient is (a_{n})(b_{m})x^{d_1+d_2}. There are n times m products.$

Method

Multiplying Polynomials Using the Box Method

Given two polynomials, their product can be calculated by using a box or table. Consider, for example, the following pair of polynomials.

P (x) Q (x) = x^{3} + 2 x^{2} - 3 = x^{2} + 4

To multiply these two polynomials, the following four steps can be followed.

Determine the Dimensions of the Table

Start by drawing a table that has as many rows as there are terms in the first polynomial and that has as many columns as there are terms in the second polynomial.

Polynomial	Number of Terms
$P (x) = x^{3} + 2 x^{2} - 3$	$3$
$Q (x) = x^{2} + 4$	$2$

For example, a table with $3$ rows and $2$ columns is needed to multiply $P (x)$ by $Q (x) .$

Write the Row and Column Labels of the Table

Now, write each term of the first polynomial at the left of each cell of the first column. Similarly, write each term of the second polynomial above each cell of the first row.

The terms x^3,2x^2, and -3 written to the left of the table. Also, the terms x^2 and 4 written above the table.

Multiply the Terms to Fill the Table

Next, fill in the table's cells by multiplying the terms written on the corresponding borders of the table. For example, the top-left cell corresponds to the product of

x^{3}

and

x^{2} .

The remaining cells can be filled by following the same procedure.

Performing the required products to fill the table.

Add Terms and Simplify

Finally, add all the expressions inside the table and combine like terms, if any.

Adding all the terms and combining like terms. The result is x^5+2x^4+4x^3+5x^2-12

The product of these polynomials has been found to be

x^{5} + 2 x^{4} + 4 x^{3} + 5 x^{2} - 12 .

Method

The FOIL Method

The FOIL method is a mnemonic for remembering how to multiply two binomials. The word FOIL is an acronym for the words

F irst,

O uter,

I nner,

and

L ast .

Consider, for example, the following product.

(x + 6) (3 x - 2)

These two binomials can be multiplied by following five steps.

Multiply the First Terms

Start by multiplying the first terms of each binomial. In this case, multiply

x

3 x .

(x + 6) (3 x - 2) = x (3 x) - 2 x + 18 x - 12

The empty box is there as a reminder that there are still missing terms.

Multiply the Outer Terms

Next, multiply the outer terms — that is, multiply the first term of the left-hand side binomial by the second term of the right-hand side binomial. In this case, multiply

x

- 2 .

(x + 6) (3 x - 2) = x (3 x) + x (- 2) 18 x - 12

Multiply the Inner Terms

Now multiply the inner terms — that is, multiply the second term of the left-hand side binomial by the first term of the right-hand side binomial. In this case, multiply

6

3 x .

(x + 6) (3 x - 2) = x (3 x) + x (- 2) + 6 (3 x) - 12

Multiply the Last Terms

Next, multiply the last terms of each binomial — that is, multiply the second term of the left-hand side binomial by the second term of the right-hand side binomial. In this case, multiply

6

- 2 .

(x + 6) (3 x - 2) = x (3 x) + x (- 2) + 6 (3 x) + 6 (- 2)

Simplify

Finally, find each product and combine like terms, if any, to simplify the resulting expression.

(x + 6) (3 x - 2) = x (3 x) + x (- 2) + 6 (3 x) + 6 (- 2)

CommutativePropMult

Commutative Property of Multiplication

(x + 6) (3 x - 2) = 3 x \cdot x + x (- 2) + 6 (3 x) + 6 (- 2)

ProdToPowTwoFac

$a \cdot a = a^{2}$

(x + 6) (3 x - 2) = 3 x^{2} + x (- 2) + 6 (3 x) + 6 (- 2)

MultPosNeg

$a (- b) = - a \cdot b$

(x + 6) (3 x - 2) = 3 x^{2} - x (2) + 6 (3 x) - 6 (2)

Multiply

(x + 6) (3 x - 2) = 3 x^{2} - 2 x + 18 x - 12

AddTerms

Add terms

(x + 6) (3 x - 2) = 3 x^{2} + 16 x - 12

The following applet illustrates the FOIL method using two arbitrary binomials.

Like any other polynomial multiplication, the FOIL method is based on the Distributive Property.

Example

Modeling the Pigpen Area

Diego's parents recently bought a piece of land where they plan to raise pigs. They need to fence off a rectangular pigpen before buying the pigs. A farmer friend told Diego that the dimensions of the pigpen, in yards, vary according to the number of pigs $x$ that are being raised in it.

A rectangular pigpen with pigs. The dimensions are (3x+1) by (5x+2).

a Write a polynomial

A (x),

in standard form, that models the pigpen area. Then, state the degree and the leading coefficient of

A (x) .

b If Diego plans to raise

15

pigs, what should be the area of the pigpen?

Hint

a The area of a rectangle equals the length times the width.

b Evaluate

A (x)

x = 15 .

Solution

a The area of a rectangle is obtained by multiplying its length by its width. Start by writing the dimensions suggested by Diego's friend.

Length : Width : 5 x + 2 3 x + 1

By multiplying the two binomials, a polynomial modeling the pigpen area will be obtained.

A (x) = (5 x + 2) (3 x + 1)

To write

A (x)

in standard form, the product on the right-hand side can be performed by using the FOIL method.

Consequently, the degree of

A (x)

2

and its leading coefficient is

15 .

b To determine the area of a pigpen large enough to raise

15

pigs, substitute

15

for

x

into

A (x) = 15 x^{2} + 11 x + 2

and simplify.

A (x) = 15 x^{2} + 11 x + 2

Substitute

$x = 15$

A (15) = 15 (15)^{2} + 11 (15) + 2

CalcPowProd

Calculate power and product

A (15) = 15 (225) + 165 + 2

Multiply

A (15) = 3375 + 165 + 2

AddTerms

Add terms

A (15) = 3542

Consequently, to raise

15

pigs, the pigpen must have an area of

3542

square yards.

Example

Determining the Area of a Triangle

Consider a right triangle on the coordinate plane with vertices

J (- 10, 0),

K (x, 0),

and

L (x, P (x)),

where

P (x) = x^{3} + 3 x^{2} - 18 x + 54 .

a Write a polynomial

A (x),

in standard form, that represents the area of

△ J K L .

Then, state the degree and leading coefficient of

A (x) .

b If

K

has coordinates

(3, 0),

what is the area of

△ J K L ?

c Which one of the following triangles has the larger area?

Hint

a The area of a triangle equals half the product of its base and its height. Note that the height of

△ J K L

P (x)

and the base is

x + 10 .

b Evaluate

A (x)

x = 3 .

c Find and compare

A (- 4)

and

A (- 1) .

Solution

a The area of a triangle is half the product of its base and its height. Since the given triangle is a right triangle, its area equals half the product of its legs.

A = \frac{1}{2} \cdot J K \cdot K L

The length of the legs can be written using the given expressions.

J K K L = x + 10 = x^{3} + 3 x^{2} - 18 x + 54

Next, substitute the previous expressions into the formula for the area.

A (x) = \frac{1}{2} (x + 10) (x^{3} + 3 x^{2} - 18 x + 54)

For simplicity, the product of the two polynomials will be computed first. Then, the resulting polynomial will be multiplied by

\frac{1}{2} .

To perform the polynomial multiplication, the Box Method can be used. To do so, start by drawing a

2 \times 4

table and writing the row and column labels.

The next step is to fill the table by multiplying the terms written on the corresponding borders.

Now, add all the terms inside the table and simplify by combining like terms.

Having computed

J K \cdot K L,

a polynomial modeling the area of

△ J K L

can be found.

A (x) = \frac{1}{2} J K \cdot K L

Substitute

$J K \cdot K L = x^{4} + 13 x^{3} + 12 x^{2} - 126 x + 540$

A (x) = \frac{1}{2} (x^{4} + 13 x^{3} + 12 x^{2} - 126 x + 540)

Distr

Distribute $\frac{1}{2}$

A (x) = \frac{1}{2} x^{4} + \frac{1 3}{2} x^{3} + 6 x^{2} - 63 x + 270

Consequently, the degree of

A (x)

4

and its leading coefficient is

\frac{1}{2} .

b If the coordinates of

K

are

(3, 0),

it implies that the coordinates of

L

are

(3, P (3)) .

To find the area of the above triangle, substitute

3

for

x

into the polynomial

A (x)

found in Part A.

A (x) = \frac{1}{2} x^{4} + \frac{1 3}{2} x^{3} + 6 x^{2} - 63 x + 270

▼

Substitute

3

for

x

and simplify

Substitute

$x = 3$

A (3) = \frac{1}{2} (3)^{4} + \frac{1 3}{2} (3)^{3} + 6 (3)^{2} - 63 (3) + 270

CalcPowProd

Calculate power and product

A (3) = \frac{8 1}{2} + \frac{3 5 1}{2} + 54 - 189 + 270

AddFrac

Add fractions

A (3) = \frac{8 1 + 3 5 1}{2} + 54 - 189 + 270

AddSubTerms

Add and subtract terms

A (3) = \frac{4 3 2}{2} + 135

CalcQuot

Calculate quotient

A (3) = 216 + 135

AddTerms

Add terms

A (3) = 351

In consequence, when

K

has coordinates

(3, 0),

the triangle

J K L

has an area of

351 .

c The triangles generated by

K_{1}

and

K_{2}

look as follows.

To determine which triangle has the larger area, find

A (- 4)

and

A (- 1)

and compare them. Start by finding

A (- 4) .

A (x) = \frac{1}{2} x^{4} + \frac{1 3}{2} x^{3} + 6 x^{2} - 63 x + 270

▼

Substitute

- 4

for

x

and simplify

Substitute

$x = - 4$

A (- 4) = \frac{1}{2} (- 4)^{4} + \frac{1 3}{2} (- 4)^{3} + 6 (- 4)^{2} - 63 (- 4) + 270

CalcPowProd

Calculate power and product

A (- 4) = \frac{2 5 6}{2} - \frac{8 3 2}{2} + 96 + 252 + 270

CalcQuot

Calculate quotient

A (- 4) = 128 - 416 + 96 + 252 + 270

AddSubTerms

Add and subtract terms

A (- 4) = 330

Next, find the value of

A (- 1) .

A (x) = \frac{1}{2} x^{4} + \frac{1 3}{2} x^{3} + 6 x^{2} - 63 x + 270

▼

Substitute

- 1

for

x

and simplify

Substitute

$x = - 1$

A (- 1) = \frac{1}{2} (- 1)^{4} + \frac{1 3}{2} (- 1)^{3} + 6 (- 1)^{2} - 63 (- 1) + 270

CalcPowProd

Calculate power and product

A (- 1) = \frac{1}{2} - \frac{1 3}{2} + 6 + 63 + 270

AddFrac

Add fractions

A (- 1) = \frac{1 - 1 3}{2} + 6 + 63 + 270

SubTerms

Subtract terms

A (- 1) = - \frac{1 2}{2} + 6 + 63 + 270

CalcQuot

Calculate quotient

A (- 1) = - 6 + 6 + 63 + 270

AddSubTerms

Add and subtract terms

A (- 1) = 333

As can be seen,

A (- 1)

is greater than

A (- 4) .

Therefore, the triangle generated by

K_{2} (- 1, 0)

has the larger area.

Example

Determining a Toy's Volume

Izabella bought her nephew a bag of magic grow toys for his birthday. Among the toys, there was a trailer and its rectangular container. Initially, the container was $5$ centimeters long, $2$ centimeters wide, and $3$ centimeters high.

A rectangular container. Dimensions: 2 cm by 5 cm by 3 cm.

The bag says that when the container is placed underwater, its dimensions will increase according to the following functions. The maximum possible size of the container is reached after being underwater for

9

hours.

ℓ (t) w (t) h (t) = \frac{1}{4} t^{2} + 5 = - \frac{1}{9} t^{2} + 2 t + 2 = t + 3

Here,

t

represents the number of hours the toy has been underwater.

a Write a polynomial

V (t),

in standard form, that represents the volume of the container after being underwater for

t

hours.

b Izabella's nephew placed the toy underwater at

10 : 00 AM

and took it out at

4 : 00 PM .

What is the volume of the container now?

Hint

a The volume of a rectangular container is obtained by multiplying its three dimensions.

b How many hours was the container underwater? Evaluate

V (t)

at this value.

Solution

a The volume of a rectangular container is found by multiplying its three dimensions.

V = w \cdot ℓ \cdot h

Therefore, to find a polynomial representing the container's volume after passing

t

hours underwater, multiply the expressions that model the dimensions of the container.

w (t) \cdot ℓ (t) \cdot h (t) ⇓ (\frac{1}{4} t^{2} + 5) (- \frac{1}{9} t^{2} + 2 t + 2) (t + 3)

By applying the Commutative Property of Multiplication, the previous product can be rewritten to have the two binomials next to each other.

(- \frac{1}{9} t^{2} + 2 t + 2) (\frac{1}{4} t^{2} + 5) (t + 3)

Then, these two binomials can be multiplied using the FOIL method. For simplicity, the trinomial will not be written during this multiplication.

Start by multiplying the first terms.
$(\frac{1}{4} t^{2} + 5) (t + 3) = \frac{1}{4} t^{2} \cdot t + \dots$
Continue by multiplying the outer terms.
$(\frac{1}{4} t^{2} + 5) (t + 3) = \frac{1}{4} t^{2} \cdot t + \frac{1}{4} t^{2} \cdot 3 + \dots$
Next, multiply the inner terms.
$(\frac{1}{4} t^{2} + 5) (t + 3) = \frac{1}{4} t^{2} \cdot t + \frac{1}{4} t^{2} \cdot 3 + 5 \cdot t + \dots$
Now, multiply the last terms.
$(\frac{1}{4} t^{2} + 5) (t + 3) = \frac{1}{4} t^{2} \cdot t + \frac{1}{4} t^{2} \cdot 3 + 5 \cdot t + 5 \cdot 3$
Finally, simplify the resulting expression by performing the required multiplications.
$(\frac{1}{4} t^{2} + 5) (t + 3) = \frac{1}{4} t^{3} + \frac{3}{4} t^{2} + 5 t + 15$

Finally, the trinomial has to be multiplied by the resulting polynomial. To perform such multiplication the Distributive Property can be used. Here, each term of the trinomial will be distributed to each term of the second polynomial.

V (t) = (- \frac{1}{9} t^{2} + 2 t + 2) (\frac{1}{4} t^{3} + \frac{3}{4} t^{2} + 5 t + 15)

▼

Multiply

(- \frac{1}{9} t^{2} + 2 t + 2)

(\frac{1}{4} t^{3} + \frac{3}{4} t^{2} + 5 t + 15)

Distr

Distribute $- \frac{1}{9} t^{2} + 2 t + 2$

V (t) = (- \frac{1}{9} t^{2} + 2 t + 2) \frac{1}{4} t^{3} + (- \frac{1}{9} t^{2} + 2 t + 2) \frac{3}{4} t^{2} + (- \frac{1}{9} t^{2} + 2 t + 2) 5 t + (- \frac{1}{9} t^{2} + 2 t + 2) 15

Distr

Distribute $\frac{1}{4} t^{3}, \frac{3}{4} t^{2}, 5 t, & 15$

V (t) = (- \frac{1}{9} t^{2} \cdot \frac{1}{4} t^{3} + 2 t \cdot \frac{1}{4} t^{3} + 2 \cdot \frac{1}{4} t^{3}) + (- \frac{1}{9} t^{2} \cdot \frac{3}{4} t^{2} + 2 t \cdot \frac{3}{4} t^{2} + 2 \cdot \frac{3}{4} t^{2}) + (- \frac{1}{9} t^{2} \cdot 5 t + 2 t \cdot 5 t + 2 \cdot 5 t) + (- \frac{1}{9} t^{2} \cdot 15 + 2 t \cdot 15 + 2 \cdot 15)

CommutativePropMult

Commutative Property of Multiplication

V (t) = (- \frac{1}{9} \cdot \frac{1}{4} t^{2} \cdot t^{3} + 2 \cdot \frac{1}{4} t \cdot t^{3} + 2 \cdot \frac{1}{4} t^{3}) + (- \frac{1}{9} \cdot \frac{3}{4} t^{2} \cdot t^{2} + 2 \cdot \frac{3}{4} t \cdot t^{2} + 2 \cdot \frac{3}{4} t^{2}) + (- \frac{1}{9} \cdot 5 t^{2} \cdot t + 2 \cdot 5 t \cdot t + 2 \cdot 5 t) + (- \frac{1}{9} \cdot 15 t^{2} + 2 \cdot 15 t + 2 \cdot 15)

▼

Simplify right-hand side

MultPow

$a^{m} \cdot a^{n} = a^{m + n}$

V (t) = (- \frac{1}{9} \cdot \frac{1}{4} t^{5} + 2 \cdot \frac{1}{4} t^{4} + 2 \cdot \frac{1}{4} t^{3}) + (- \frac{1}{9} \cdot \frac{3}{4} t^{4} + 2 \cdot \frac{3}{4} t^{3} + 2 \cdot \frac{3}{4} t^{2}) + (- \frac{1}{9} \cdot 5 t^{3} + 2 \cdot 5 t^{2} + 2 \cdot 5 t) + (- \frac{1}{9} \cdot 15 t^{2} + 2 \cdot 15 t + 2 \cdot 15)

MultFrac

Multiply fractions

V (t) = (- \frac{1}{3 6} t^{5} + \frac{2}{4} t^{4} + \frac{2}{4} t^{3}) + (- \frac{3}{3 6} t^{4} + \frac{6}{4} t^{3} + \frac{6}{4} t^{2}) + (- \frac{5}{9} t^{3} + 2 \cdot 5 t^{2} + 2 \cdot 5 t) + (- \frac{1 5}{9} t^{2} + 2 \cdot 15 t + 2 \cdot 15)

SimpQuotProd

Simplify quotient and product

V (t) = (- \frac{1}{3 6} t^{5} + \frac{1}{2} t^{4} + \frac{1}{2} t^{3}) + (- \frac{1}{1 2} t^{4} + \frac{3}{2} t^{3} + \frac{3}{2} t^{2}) + (- \frac{5}{9} t^{3} + 10 t^{2} + 10 t) + (- \frac{5}{3} t^{2} + 30 t + 30)

CommutativePropAdd

Commutative Property of Addition

V (t) = - \frac{1}{3 6} t^{5} + (\frac{1}{2} t^{4} - \frac{1}{1 2} t^{4}) + (\frac{1}{2} t^{3} - \frac{5}{9} t^{3} + \frac{3}{2} t^{3}) + (\frac{3}{2} t^{2} - \frac{5}{3} t^{2} + 10 t^{2}) + (10 t + 30 t) + 30

AddSubTerms

Add and subtract terms

V (t) = - \frac{1}{3 6} t^{5} + \frac{5}{1 2} t^{4} + \frac{1 3}{9} t^{3} + \frac{5 9}{6} t^{2} + 40 t + 30

b The polynomial

V (t)

found in Part A gives the volume of the container after passing

t

hours underwater. It is given that Izabella's nephew put the container underwater at

10 : 00 AM

and took it out at

4 : 00 PM .

Therefore, the container was underwater for

6

hours.

6 hours 10 : 00 AM ⟶ 4 : 00 PM

Consequently, to find the current volume of the container, evaluate

V (t)

t = 6 .

V (t) = - \frac{1}{3 6} t^{5} + \frac{5}{1 2} t^{4} + \frac{1 3}{9} t^{3} + \frac{5 9}{6} t^{2} + 40 t + 30

▼

Substitute

6

for

t

and evaluate

Substitute

$t = 6$

V (6) = - \frac{1}{3 6} (6)^{5} + \frac{5}{1 2} (6)^{4} + \frac{1 3}{9} (6)^{3} + \frac{5 9}{6} (6)^{2} + 40 (6) + 30

CalcPow

Calculate power

V (6) = - \frac{1}{3 6} \cdot 7776 + \frac{5}{1 2} \cdot 1296 + \frac{1 3}{9} \cdot 216 + \frac{5 9}{6} \cdot 36 + 40 (6) + 30

MoveRightFacToNum

$\frac{a}{c} \cdot b = \frac{a \cdot b}{c}$

V (6) = - \frac{7 7 7 6}{3 6} + \frac{5 \cdot 1 2 9 6}{1 2} + \frac{1 3 \cdot 2 1 6}{9} + \frac{5 9 \cdot 3 6}{6} + 40 (6) + 30

Multiply

V (6) = - \frac{7 7 7 6}{3 6} + \frac{6 4 8 0}{1 2} + \frac{2 8 0 8}{9} + \frac{2 1 2 4}{6} + 240 + 30

CalcQuot

Calculate quotient

V (6) = - 216 + 540 + 312 + 354 + 240 + 30

AddSubTerms

Add and subtract terms

V (6) = 1260

After passing

6

hours underwater, the container will have a volume of

1260

cubic centimeters.

Discussion

Analyzing the Product of Two Polynomials

In all the examples seen throughout this lesson, the product of two or more polynomials has resulted in a new polynomial. This is not a coincidence. In fact, the following property guarantees that multiplying polynomials always produces a polynomial.

Rule

Closure Property of Polynomial Multiplication

Given two polynomials $P (x)$ and $Q (x),$ the product $P (x) \cdot Q (x)$ is always a polynomial.

Multiplying two polynomials produces a new polynomial.

In other words, the polynomials are closed under multiplication.

Proof

Consider two arbitrary polynomials

P (x)

and

Q (x)

written in standard form.

P (x) Q (x) = a_{n} x^{d_{1}} + \dots + a_{1} x + a_{0} = b_{n} x^{d_{2}} + \dots + b_{1} x + b_{0}

These two polynomials can be multiplied by using the Distributive Property. The Product of Powers Property can also be applied to simplify the resulting expression.

Since

P (x)

and

Q (x)

are polynomials, all the exponents are whole numbers. Furthermore, because the whole numbers are closed under addition, the exponents of the resulting expression are whole numbers. Then, the new expression can be rewritten as follows.

P (x) Q (x) = c_{k} x^{D} + \dots + c_{1} x + c_{0}

Consequently, the new expression is a polynomial. Therefore, the product of two polynomials produces a polynomial, which proves that the polynomials are closed under multiplication.

Pop Quiz

Practicing Multiplication of Polynomials

Given two polynomials $P (x)$ and $Q (x),$ compute their product and find the required information.

Two random polynomials P(x) and Q(x) are given. A certain information about P(x)Q(x) is required (the degree, leading coefficient, constant term, linear term, quadratic term, cubic term, quartic term).

Closure

Computing a Polynomials Product

Before finishing this lesson, take another look at the challenge presented at the beginning. There it was given the following pair of polynomials.

P (x) Q (x) = 2 x^{3} - x + 5 = 4 (x + 2) (x - 1)

In the challenge, it was asked whether it was possible to find the product of

P (x)

and

Q (x) .

Now, with all the knowledge learned, it can be said that it is possible to multiply these two polynomials, and the resulting expression will also be a polynomial. To compute

P (x) \cdot Q (x),

first, rewrite

Q (x)

in standard form. To do so, the FOIL Method can be applied.

Q (x) = 4 (x + 2) (x - 1)

MultII

Multiply $(x + 2)$ by $(x - 1)$

Q (x) = 4 (x (x) + x (- 1) + 2 x + 2 (- 1))

▼

Simplify right-hand side

MultPow

$a^{m} \cdot a^{n} = a^{m + n}$

Q (x) = 4 (x^{2} + x (- 1) + 2 x + 2 (- 1))

MultPosNeg

$a (- b) = - a \cdot b$

Q (x) = 4 (x^{2} - x \cdot 1 + 2 x - 2 \cdot 1)

MultByOne

$a \cdot 1 = a$

Q (x) = 4 (x^{2} - x + 2 x - 2)

AddTerms

Add terms

Q (x) = 4 (x^{2} + x - 2)

Distr

Distribute $4$

Q (x) = 4 x^{2} + 4 x - 8

Having both polynomials written in standard form, the required product can be found by using the Distributive Property.

P (x) \cdot Q (x)

SubstituteExpressions

Substitute expressions

(2 x^{3} - x + 5) (4 x^{2} + 4 x - 8)

▼

Distribute

2 x^{3} - x + 5

Distr

Distribute $2 x^{3} - x + 5$

(2 x^{3} - x + 5) 4 x^{2} + (2 x^{3} - x + 5) 4 x + (2 x^{3} - x + 5) (- 8)

Distr

Distribute $4 x^{2}$

(2 x^{3} \cdot 4 x^{2} - x \cdot 4 x^{2} + 5 \cdot 4 x^{2}) + (2 x^{3} - x + 5) 4 x + (2 x^{3} - x + 5) (- 8)

Distr

Distribute $4 x$

(2 x^{3} \cdot 4 x^{2} - x \cdot 4 x^{2} + 5 \cdot 4 x^{2}) + (2 x^{3} \cdot 4 x - x \cdot 4 x + 5 \cdot 4 x) + (2 x^{3} - x + 5) (- 8)

Distr

Distribute $- 8$

(2 x^{3} \cdot 4 x^{2} - x \cdot 4 x^{2} + 5 \cdot 4 x^{2}) + (2 x^{3} \cdot 4 x - x \cdot 4 x + 5 \cdot 4 x) + (2 x^{3} (- 8) - x (- 8) + 5 (- 8))

▼

Simplify

MultPow

$a^{m} \cdot a^{n} = a^{m + n}$

(2 \cdot 4 x^{5} - 4 x^{3} + 5 \cdot 4 x^{2}) + (2 \cdot 4 x^{4} - 4 x^{2} + 5 \cdot 4 x) + (2 x^{3} (- 8) - x (- 8) + 5 (- 8))

Multiply

(8 x^{5} - 4 x^{3} + 20 x^{2}) + (8 x^{4} - 4 x^{2} + 20 x) + (- 16 x^{3} + 8 x - 40)

CommutativePropAdd

Commutative Property of Addition

8 x^{5} + 8 x^{4} + (- 4 x^{3} - 16 x^{3}) + (20 x^{2} - 4 x^{2}) + (20 x + 8 x) - 40

AddSubTerms

Add and subtract terms

8 x^{5} + 8 x^{4} - 20 x^{3} + 16 x^{2} + 28 x - 40

As expected, the resulting expression is a polynomial. The degree of the resulting polynomial is

5,

its leading coefficient is

8,

and its constant term is

- 40 .

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Catch-Up and Review

Hint

Solution

Hint

Solution

Hint

Solution

Hint

Solution

Proof