5.6 Divide Polynomials

Learning Objectives

By the end of this section, you will be able to:

  • Divide a polynomial by a monomial
  • Divide a polynomial by a binomial

Try It

Before you get started, take this readiness quiz:

1) Add: [latex]\frac{3}{d}+\frac{x}{d}[/latex]
2) Simplify: [latex]\frac{30x{y}^{3}}{5xy}[/latex]
3) Combine like terms: [latex]8{a}^{2}+12a+1+3{a}^{2}-5a+4[/latex]

Divide a Polynomial by a Monomial

In the last section, you learned how to divide a monomial by a monomial. As you continue to build up your knowledge of polynomials the next procedure is to divide a polynomial of two or more terms by a monomial.

The method we’ll use to divide a polynomial by a monomial is based on the properties of fraction addition. So we’ll start with an example to review fraction addition.

[latex]\displaystyle\frac{y}{5}+\frac{2}{5}[/latex]

The sum simplifies to

[latex]\displaystyle\frac{y+2}{5}[/latex]

Now we will do this in reverse to split a single fraction into separate fractions.

We’ll state the fraction addition property here just as you learned it and in reverse

Fraction Addition

If [latex]a[/latex], [latex]b[/latex] and [latex]c[/latex] are numbers where [latex]c\neq{0}[/latex], then

[latex]\displaystyle\frac ac+\frac bc=\frac{a+b}c[/latex] and [latex]\displaystyle\frac{a+b}c =\frac ac+\frac bc[/latex]

We use the form on the left to add fractions and we use the form on the right to divide a polynomial by a monomial.

For example, [latex]\frac{y+2}{5}[/latex] can be written [latex]\frac{y}{5}+\frac{2}{5}[/latex]

We use this form of fraction addition to divide polynomials by monomials.

Division of a Polynomial by a Monomial

To divide a polynomial by a monomial, divide each term of the polynomial by the monomial.

Example 5.6.1

Find the quotient: [latex]\frac{7{y}^{2}+21}{7}[/latex]

Solution

Step 1: Divide each term of the numerator by the denominator.

[latex]\frac{7{y}^{2}}{7}+\frac{21}{7}[/latex]

Step 2: Simplify each fraction.

[latex]{y}^{2}+3[/latex]

Try It

4) Find the quotient: [latex]\frac{8{z}^{2}+24}{4}[/latex]

Solution

[latex]2{z}^{2}+6[/latex]

5) Find the quotient: [latex]\frac{18{z}^{2}-27}{9}[/latex]

Solution

[latex]2{z}^{2}-3[/latex]

Remember that division can be represented as a fraction. When you are asked to divide a polynomial by a monomial and it is not already in fraction form, write a fraction with the polynomial in the numerator and the monomial in the denominator.

Example 5.6.2

Find the quotient: [latex](18{x}^{3}-36{x}^{2})\div{6x}[/latex]

Solution

Step 1: Rewrite as a fraction.

[latex]\frac{18{x}^{3}-36{x}^{2}}{6x}[/latex]

Step 2: Divide each term of the numerator by the denominator.

[latex]\begin{eqnarray*}&=&\frac{18{x}^{3}}{6x}-\frac{36{x}^{2}}{6x}\\[1ex]\text{Simplify.}\;\;&=&3{x}^{2}-6x\end{eqnarray*}[/latex]

Try It

6) Find the quotient: [latex](27{b}^{3}-33{b}^{2})\div{3b}[/latex]

Solution

[latex]9{b}^{2}-11b[/latex]

7) Find the quotient: [latex](25{y}^{3}-55{y}^{2})\div{5y}[/latex]

Solution

[latex]5{y}^{2}-11y[/latex]

When we divide by a negative, we must be extra careful with the signs.

Example 5.6.3

Find the quotient: [latex]\frac{12{d}^{2}-16d}{-4}[/latex]

Solution

Step 1: Divide each term of the numerator by the denominator.

[latex]\frac{12{d}^{2}}{-4}-\frac{16d}{-4}[/latex]

Step 2: Simplify. Remember, subtracting a negative is like adding a positive!

[latex]-3{d}^{2}+4d[/latex]

Try It

8) Find the quotient: [latex]\frac{25{y}^{2}-15y}{-5}[/latex]

Solution

[latex]-5{y}^{2}+3y[/latex]

9) Find the quotient: [latex]\frac{42{b}^{2}-18b}{-6}[/latex]

Solution

[latex]-7{b}^{2}+3b[/latex]

Example 5.6.4

Find the quotient: [latex]\frac{105{y}^{5}+75{y}^{3}}{5{y}^{2}}[/latex]

Solution

Step 1: Separate the terms.

[latex]\begin{eqnarray*}&=&\frac{105{y}^{5}}{5{y}^{2}}+\frac{75{y}^{3}}{5{y}^{2}}\\[1ex]\text{Simplify.}\;\;&=&21{y}^{3}+15y\end{eqnarray*}[/latex]

Try It

10) Find the quotient: [latex]\frac{60{d}^{7}+24{d}^{5}}{4{d}^{3}}[/latex]

Solution

[latex]15{d}^{4}+6{d}^{2}[/latex]

11) Find the quotient: [latex]\frac{216{p}^{7}-48{p}^{5}}{6{p}^{3}}[/latex]

Solution

[latex]36{p}^{4}-8{p}^{2}[/latex]

Example 5.6.5

Find the quotient: [latex](15{x}^{3}y-35x{y}^{2})\div(-5xy)[/latex]

Solution

Step 1: Rewrite as a fraction.

[latex]\frac{15{x}^{3}y-35x{y}^{2}}{-5xy}[/latex]

Step 2: Separate the terms.

[latex]\begin{eqnarray*}&=&\frac{15{x}^{3}y}{-5xy}-\frac{35x{y}^{2}}{-5xy}\\[1ex]\text{Simplify.}\;\;&=&-3{x}^{2}+7y\end{eqnarray*}[/latex]

Try It

12) Find the quotient: [latex](32{a}^{2}b-16a{b}^{2})\div(-8ab)[/latex]

Solution

[latex]-4a+2b[/latex]

13) Find the quotient: [latex](-48{a}^{8}{b}^{4}-36{a}^{6}{b}^{5})\div(-6{a}^{3}{b}^{3})[/latex]

Solution

[latex]8{a}^{5}b+6{a}^{3}{b}^{2}[/latex]

Example 5.6.6

Find the quotient: [latex]\frac{36x^3y^2+27x^2y^2-9x^2y^3}{9x^2y}[/latex]
Solution

Step 1: Separate the terms.

[latex]\begin{eqnarray*}&=&\frac{36{x}^{3}{y}^{2}}{9{x}^{2}y}+\frac{27{x}^{2}{y}^{2}}{9{x}^{2}y}-\frac{9{x}^{2}{y}^{3}}{9{x}^{2}y}\\[1ex]\text{Simplify.}\;\;&=&4xy+3y-{y}^{2}\end{eqnarray*}[/latex]

Try It

14) Find the quotient: [latex]\frac{40{x}^{3}{y}^{2}+24{x}^{2}{y}^{2}-16{x}^{2}{y}^{3}}{8{x}^{2}y}[/latex].
Solution
[latex]5xy+3y-2{y}^{2}[/latex]

15) Find the quotient: [latex]\frac{35{a}^{4}{b}^{2}+14{a}^{4}{b}^{3}-42{a}^{2}{b}^{4}}{7{a}^{2}{b}^{2}}[/latex]

Solution
[latex]5{a}^{2}+2{a}^{2}b-6{b}^{2}[/latex]

Example 5.6.7

Find the quotient: [latex]\frac{10{x}^{2}+5x-20}{5x}[/latex]

Solution

Step 1: Separate the terms.

[latex]\begin{eqnarray*}&=&\frac{10{x}^{2}}{5x}+\frac{5x}{5x}-\frac{20}{5x}\\[1ex]\text{Simplify.}\;\;&=&2x+1+\frac{4}{x}\end{eqnarray*}[/latex]

Try It

16) Find the quotient: [latex]\frac{18{c}^{2}+6c-9}{6c}[/latex]

Solution

[latex]3c+1-\frac{3}{2c}[/latex]

17) Find the quotient: [latex]\frac{10{d}^{2}-5d-2}{5d}[/latex]

Solution

[latex]2d-1-\frac{2}{5d}[/latex]

Divide a Polynomial by a Binomial

To divide a polynomial by a binomial, we follow a procedure very similar to long division of numbers. So let’s look carefully the steps we take when we divide a 3-digit number, [latex]875[/latex], by a 2-digit number, [latex]25[/latex].

We write the long division
The long division of 875 by 25.
Figure 5.6.1
We divide the first two digits, [latex]87[/latex], by [latex]25[/latex].
25 fits into 87 three times. 3 is written above the second digit of 875 in the long division bracket.
Figure 5.6.2
We multiply [latex]3[/latex] times [latex]25[/latex] and write the product under the [latex]87[/latex].
The product of 3 and 25 is 75, which is written below the first two digits of 875 in the long division bracket.
Figure 5.6.3
Now we subtract [latex]75[/latex] from [latex]87[/latex].
87 minus 75 is 12, which is written under 75.
Figure 5.6.4
Then we bring down the third digit of the dividend, [latex]5[/latex].
The 5 in 875 is brought down next to the 12, making 125.
Figure 5.6.5
Repeat the process, dividing [latex]25[/latex] into [latex]125[/latex].
25 fits into 125 five times. 5 is written to the right of the 3 on top of the long division bracket. 5 times 25 is 125. 125 minus 125 is zero. There is zero remainder, so 25 fits into 125 exactly five times. 875 divided by 25 equals 35.
Figure 5.6.6

We check division by multiplying the quotient by the divisor.

If we did the division correctly, the product should equal the dividend.

[latex]35\times{25}\\875\checkmark[/latex]

Now we will divide a trinomial by a binomial. As you read through the example, notice how similar the steps are to the numerical example above.

Example 5.6.8

Find the quotient: [latex]({x}^{2}+9x+20)\div(x+5)[/latex]

Solution

Step 1: Write it as a long division problem.
Step 2: Be sure the dividend is in standard form.

The long division of x squared plus 9 x plus 20 by x plus 5
Figure 5.6.7

Step 3: Divide [latex]x^2[/latex] by [latex]x[/latex].
It may help to ask yourself, “What do I need to multiply [latex]x[/latex] by to get [latex]x^2[/latex]?”
Step 4: Put the answer, [latex]x[/latex], in the quotient over the [latex]x[/latex] term.

x fits into x squared x times. x is written above the second term of x squared plus 9 x plus 20 in the long division bracket.
Figure 5.6.8

Step 5: Multiply [latex]x[/latex] times [latex]x + 5[/latex].
Line up the like terms under the dividend.
Step 6: Subtract [latex]x^2+5x[/latex] from [latex]x^2+9x[/latex].

The product of x and x plus 5 is x squared plus 5 x, which is written below the first two terms of x squared plus 9x plus 20 in the long division bracket.
Figure 5.6.9

Step 7: Add, changing the signs might make it easier to do this. 
Step 8: Then bring down the last term, [latex]20[/latex].

The sum of x squared plus 9 x and negative x squared plus negative 5 x is 4 x, which is written underneath the negative 5 x. The third term in x squared plus 9 x plus 20 is brought down next to 4 x, making 4 x plus 20.
Figure 5.6.10

Step 9: Divide [latex]4x[/latex] by [latex]x[/latex].
It may help to ask yourself, “What do I need to multiply [latex]x[/latex] by to get [latex]4x[/latex]?”
Step 10: Put the answer, [latex]4[/latex], in the quotient over the constant term.

4 x divided by x is 4. Plus 4 is written on top of the long division bracket, next to x and above the 20 in x squared plus 9 x plus 20.
Figure 5.6.11

Step 11: Multiply [latex]4[/latex] times [latex]x+5[/latex].

x plus 5 times 4 is 4 x plus 20, which is written under the first 4 x plus 20.
Figure 5.6.12

Step 12: Subtract [latex]4x+20[/latex] from [latex]4x+20[/latex].

4 x plus 20 minus 4 x plus 20 is 0. The remainder is 0. x squared plus 9 x plus 20 divided by x plus 5 equals x plus 4.
Figure 5.6.13

Step 13: Check:
Multiply the quotient by the divisor.
You should get the dividend.

[latex](x+4)(x+5)\\x^2+9x+20\checkmark[/latex]

Try It

18) Find the quotient: [latex]({y}^{2}+10y+21)\div(y+3)[/latex]

Solution

[latex]y+7[/latex]

19) Find the quotient: [latex]({m}^{2}+9m+20)\div(m+4)[/latex]

Solution

[latex]m+5[/latex]

When the divisor has a subtraction sign, we must be extra careful when we multiply the partial quotient and then subtract. It may be safer to show that we change the signs and then add.

Example 5.6.9

Find the quotient: ([latex]2{x}^{2}-5x-3)\div(x-3)[/latex]

Solution

Step 1: Write it as a long division problem.
Step 2: Be sure the dividend is in standard form.

The long division of 2 x squared minus 5 x minus 3 by x minus 3.
Figure 5.6.14

Step 3: Divide [latex]2x^2[/latex] by [latex]x[/latex].
Step 4: Put the answer, [latex]2x[/latex], in the quotient over the [latex]x[/latex] term.

x fits into 2 x squared 2 x times. 2 x is written above the second term of 2 x squared minus 5 x minus 3 in the long division bracket.
Figure 5.6.15

Step 5: Multiply [latex]2x[/latex] times [latex]x-3[/latex].
Line up the like terms under the dividend.

The product of 2 x and x minus 3 is 2 x squared minus 6 x, which is written below the first two terms of 2 x squared minus 5 x minus 3 in the long division bracket.
Figure 5.6.16

Step 6: Subtract [latex]2x^2-6x[/latex] from [latex]2x^2-5x[/latex].
Step 7: Change the signs and then add.
Step 8: Then bring down the last term.

The sum of 2 x squared minus 5 x and negative 2 x squared plus 6 x is x, which is written underneath the 6 x. The third term in 2 x squared minus 5 x minus 3 is brought down next to x, making x minus 3.
Figure 5.6.17

Step 9: Divide [latex]x[/latex] by [latex]x[/latex].
Step 10: Put the answer, [latex]1[/latex], in the quotient over the constant term.

Plus 1 is written on top of the long division bracket, next to 2 x and above the minus 3 in 2 x squared minus 5 x minus 3.
Figure 5.6.18

Step 11: Multiply [latex]1[/latex] times [latex]x-3[/latex].

x minus 3 times 1 is x minus 3, which is written under the first x minus 3.
Figure 5.6.19

Step 12: Subtract [latex]x-3[/latex] from [latex]x-3[/latex] by changing the signs and adding.

The binomial x minus 3 minus the binomial negative x plus 3 is 0. The remainder is 0. 2 x squared minus 5 x minus 3 divided by x minus 3 equals 2 x plus 1.
Figure 5.6.20

Step 13: To check, multiply [latex](x-3)(2x+1)[/latex].

The result should be [latex]2x^2-5x-3[/latex].

Try It

20) Find the quotient: [latex](2{x}^{2}-3x-20)\div(x-4)[/latex]

Solution

[latex]2x+5[/latex]

21) Find the quotient: [latex](3{x}^{2}-16x-12)\div(x-6)[/latex]

Solution

[latex]3x+2[/latex]

When we divided [latex]875[/latex] by [latex]25[/latex], we had no remainder. But sometimes division of numbers does leave a remainder. The same is true when we divide polynomials. In example 5.6.10 we’ll have a division that leaves a remainder. We write the remainder as a fraction with the divisor as the denominator.

Example 5.6.10

Find the quotient: [latex]({x}^{3}-{x}^{2}+x+4)\div(x+1)[/latex]

Solution

Step 1: Write it as a long division problem.
Step 2: Be sure the dividend is in standard form.

The long division of x cubed minus x squared plus x plus 4 by x plus 1.
Figure 5.6.21

Step 3: Divide [latex]x^3[/latex] by [latex]x[/latex].
Step 4: Put the answer, [latex]x^2[/latex], in the quotient over the [latex]x^2[/latex] term.
Step 5: Multiply [latex]x^2[/latex] times [latex]x+1[/latex].
Line up the like terms under the dividend.

x fits into x squared x times. x is written above the second term of x cubed minus x squared plus x plus 4 in the long division bracket.
Figure 5.6.22

Step 6: Subtract [latex]x^3+x^2[/latex] from [latex]x^3-x^2[/latex] by changing the signs and adding.
Step 7: Then bring down the next term.

The sum of x cubed minus x squared and negative x cubed plus negative x squared is negative 2 x squared, which is written underneath the negative x squared. The next term in x cubed minus x squared plus x plus 4 is brought down next to negative 2 x squared, making negative 2 x squared plus x.
Figure 5.6.23

Step 9: Divide [latex]-2x^2[/latex] by [latex]x[/latex].
Step 10: Put the answer, [latex]-2x[/latex], in the quotient over the [latex]x[/latex] term.
Step 11: Multiply [latex]-2x[/latex] times [latex]x+1[/latex].
Line up the like terms under the dividend.

Minus 2 x is written on top of the long division bracket, next to x squared and above the x in x cubed minus x squared plus x plus 4. Negative 2 x squared minus 2 x is written under negative 2 x squared plus x.
Figure 5.6.24

Step 12: Subtract [latex]-2x^2-2x[/latex] from [latex]-2x^2+x[/latex] by changing the signs and adding.
Step 13: Then bring down the last term.

The sum of negative 2 x squared plus x and 2 x squared plus 2 x is found to be 3 x. The last term in x cubed minus x squared plus x plus 4 is brought down, making 3 x plus 4.
Figure 5.6.25

Step 14: Divide [latex]3x[/latex] by [latex]x[/latex].
Step 15: Put the answer, [latex]3[/latex], in the quotient over the constant term.
Step 16: Multiply [latex]3[/latex] times [latex]x+1[/latex]. Line up the like terms under the dividend.

Plus 3 is written on top of the long division bracket, above the 4 in x cubed minus x squared plus x plus 4. 3 x plus 3 is written under 3 x plus 4.
Figure 5.6.26

Step 17: Subtract [latex]3x+3[/latex] from [latex]3x+4[/latex] by changing the signs and adding.
Step 18: Write the remainder as a fraction with the divisor as the denominator.

The sum of 3 x plus 4 and negative 3 x plus negative 3 is 1. Therefore, the polynomial x cubed minus x squared plus x plus 4, divided by the binomial x plus 1, equals x squared minus 2 x plus the fraction 1 over x plus 1.
Figure 5.6.27

Step 19: To check, multiply [latex](x+1)({x}^{2}-2x+3+\frac{1}{x+1})[/latex].

The result should be [latex]{x}^{3}-{x}^{2}+x+4[/latex]

Try It

22) Find the quotient: [latex]({x}^{3}+5{x}^{2}+8x+6)\div(x+2)[/latex]

Solution

[latex]\left({x}^{2}+3x+2+\frac{2}{x+2}\right)[/latex]

23) Find the quotient: [latex](2{x}^{3}+8{x}^{2}+x-8)\div(x+1)[/latex]

Solution

[latex]\left(2{x}^{2}+6x-5-\frac{3}{x+1}\right)[/latex]

Look back at the dividends in  Examples 5.6.8, 5.6.9, and 5.6.10 The terms were written in descending order of degrees, and there were no missing degrees. The dividend in Example 5.6.11 will be [latex]{x}^{4}-{x}^{2}+5x-2[/latex]. It is missing an [latex]{x}^{3}[/latex] term. We will add in [latex]0{x}^{3}[/latex] as a placeholder.

Example 5.6.11

Find the quotient: [latex]({x}^{4}-{x}^{2}+5x-2)\div(x+2)[/latex]

Solution

Notice that there is no [latex]{x}^{3}[/latex] term in the dividend. We will add [latex]0{x}^{3}[/latex] as a placeholder.

Step 1: Write it as a long division problem.
Be sure the dividend is in standard form with placeholders for missing terms.

The long division of x to the fourth power plus 0 x cubed minus x squared minus 5 x minus 2 by x plus 2.
Figure 5.6.28

Step 2: Divide [latex]x^4[/latex] by [latex]x[/latex].
Step 3: Put the answer, [latex]x^3[/latex], in the quotient over the [latex]x^3[/latex] term.
Step 4: Multiply [latex]x^3[/latex] times [latex]x+2[/latex].
Line up the like terms.
Step 5: Subtract and then bring down the next term.

x cubed is written on top of the long division bracket above the x cubed term in the dividend. Below the first two terms of the dividend x to the fourth power plus 2 x cubed is subtracted to give negative 2 x cubed minus x squared. A note next to the division reads “It may be helpful to change the signs and add.”
Figure 5.6.29

Step 6: Divide [latex]-2x^3[/latex] by [latex]x[/latex].
Step 7: Put the answer, [latex]-2x^2[/latex], in the quotient over the [latex]x^2[/latex] term.
Step 8: Multiply [latex]-2x^2[/latex] times [latex]x+1[/latex].
Line up the like terms.
Step 9: Subtract and bring down the next term.

x cubed minus 2 x squared is written on top of the long division bracket. At the bottom of the long division negative 2 x cubed minus 4 x squared is subtracted to give 3 x squared plus 5 x. A note reads “It may be helpful to change the signs and add.”
Figure 5.6.30

Step 10: Divide [latex]3x^2[/latex] by [latex]x[/latex].
Step 11: Put the answer, [latex]3x[/latex], in the quotient over the [latex]x[/latex] term.
Step 12: Multiply [latex]3x[/latex] times [latex]x+1[/latex].
Line up the like terms.
Step 13: Subtract and bring down the next term.

x cubed minus 2 x squared plus 3 x is written on top of the long division bracket. At the bottom of the long division 3 x squared plus 6 x is subtracted to give negative x minus 2. A note reads “It may be helpful to change the signs and add.”
Figure 5.6.31

Step 14: Divide [latex]-x[/latex] by [latex]x[/latex].
Step 15: Put the answer, [latex]-1[/latex], in the quotient over the constant term.
Step 16: Multiply [latex]-1[/latex] times [latex]x+1[/latex].
Line up the like terms.
Step 17: Change the signs, add.

x cubed minus 2 x squared plus 3 x minus 1 is written on top of the long division bracket. At the bottom of the long division negative x minus 2 is subtract to give 0. A note reads “It may be helpful to change the signs and add.” The polynomial x to the fourth power minus x squared plus 5 x minus 2, divided by the binomial x plus 2 equals the polynomial x cubed minus 2 x squared plus 3 x minus 1.
Figure 5.6.32

Step 18: To check, multiply [latex](x+2)({x}^{3}-2{x}^{2}+3x-1)[/latex].

The result should be [latex]({x}^{4}-{x}^{2}+5x-2)[/latex]

Try It

24) Find the quotient: [latex]({x}^{3}+3x+14)\div(x+2)[/latex]

Solution

[latex]{x}^{2}-2x+7[/latex]

25) Find the quotient: [latex]({x}^{4}-3{x}^{3}-1000)\div(x+5)[/latex]

Solution

[latex]{x}^{3}-8{x}^{2}+40x-200[/latex]

In Example 5.6.12, we will divide by [latex]2a-3[/latex]. As we divide we will have to consider the constants as well as the variables.

Example 5.6.12

Find the quotient: [latex](8{a}^{3}+27)\div(2a+3)[/latex]

Solution

This time we will show the division all in one step. We need to add two placeholders in order to divide.

The figure shows the long division of 8 a cubed plus 27 by 2 a plus 3. In the long division bracket, placeholders 0 a squared and 0 a are added into the polynomial. On the first line under the dividend 8 a cubed plus 12 a squared is subtracted. To the right, an arrow indicates that this value came from multiplying 4 a squared by 2 a plus 3. The subtraction gives negative 12 a squared plus 0 a. From this negative 12 a squared minus 18 a is subtracted. To the right, an arrow indicates that this value came from multiplying 6 a by 2 a plus 3. The subtraction give 18 a plus 27. From this 18 a plus 27 is subtracted. To the right, an arrow indicates that this value came from multiplying 9 by 2 a plus 3. The result is 0.
Figure 5.6.33

To check, multiply [latex](2a+3)(4{a}^{2}-6a+9)[/latex]

The result should be [latex]8{a}^{3}+27[/latex]

Try It

26) Find the quotient: [latex]({x}^{3}-64)\div(x-4)[/latex]

Solution

[latex]{x}^{2}+4x+16[/latex]

27) Find the quotient: [latex](125{x}^{3}-8)\div(5x-2)[/latex]

Solution

[latex]25{x}^{2}+10x+4[/latex]

Access these online resources for additional instruction and practice with dividing polynomials:

Key Concepts

  • Fraction Addition
    • If [latex]a[/latex], [latex]b[/latex] and [latex]c[/latex] are numbers where [latex]c\neq{0}[/latex], then
      [latex]\displaystyle\frac{a}{c}+\frac{b}{c}=\frac{a+b}{c}[/latex] and [latex]\displaystyle\frac{a+b}{c}=\frac{a}{c}+\frac{b}{c}[/latex]

  • Division of a Polynomial by a Monomial
    • To divide a polynomial by a monomial, divide each term of the polynomial by the monomial.

Self Check

After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

After reviewing this checklist, what will you do to become confident for all goals?

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