Dividing Polynomials

5.4 Dividing Polynomials

Learning Objectives

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

  • Dividing monomials
  • Dividing a polynomial by a monomial
  • Dividing polynomials using long division
  • Dividing polynomials using synthetic division
  • Dividing polynomial functions
  • Use the remainder and factor theorems

Be Prepared 5.10

Before you get started, take this readiness quiz.

Add: 3d+xd.3d+xd.
If you missed this problem, review Example 1.28.

Be Prepared 5.11

Simplify: 30xy35xy.30xy35xy.
If you missed this problem, review Example 1.25.

Be Prepared 5.12

Combine like terms: 8a2+12a+1+3a2−5a+4.8a2+12a+1+3a2−5a+4.
If you missed this problem, review Example 1.7.

Dividing Monomials

We are now familiar with all the properties of exponents and used them to multiply polynomials. Next, we’ll use these properties to divide monomials and polynomials.

Example 5.36

Find the quotient: 54a2b3÷(−6ab5).54a2b3÷(−6ab5).

Solution

When we divide monomials with more than one variable, we write one fraction for each variable.

54a2b3÷(−6ab5)54a2b3÷(−6ab5)
Rewrite as a fraction. 54a2b3−6ab554a2b3−6ab5
Use fraction multiplication. 54−6·a2a·b3b554−6·a2a·b3b5
Simplify and use the Quotient Property. −9·a·1b2−9·a·1b2
Multiply. −9ab2−9ab2
This table illustrates the step-by-step simplification of an algebraic expression involving division.

Try It 5.71

Find the quotient: −72a7b3÷(8a12b4).−72a7b3÷(8a12b4).

Try It 5.72

Find the quotient: −63c8d3÷(7c12d2).−63c8d3÷(7c12d2).

Once you become familiar with the process and have practiced it step by step several times, you may be able to simplify a fraction in one step.

Example 5.37

Find the quotient: 14x7y1221x11y6.14x7y1221x11y6.

Solution

Be very careful to simplify 14211421 by dividing out a common factor, and to simplify the variables by subtracting their exponents.

14x7y1221x11y614x7y1221x11y6
Simplify and use the Quotient Property. 2y63x42y63x4
This table illustrates the simplification of a rational algebraic expression, showing the original expression, the instruction to simplify using the Quotient Property, and the resulting simplified expression.

Try It 5.73

Find the quotient: 28x5y1449x9y12.28x5y1449x9y12.

Try It 5.74

Find the quotient: 30m5n1148m10n14.30m5n1148m10n14.

Divide a Polynomial by a Monomial

Now that we know how to divide a monomial by a monomial, 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. The sum y5+25y5+25 simplifies to y+25.y+25.

Now we will do this in reverse to split a single fraction into separate fractions. For example, y+25y+25 can be written y5+25.y5+25.

This is the “reverse” of fraction addition and it states that if a, b, and c are numbers where c≠0,c≠0, then a+bc=ac+bc.a+bc=ac+bc. We will use this 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 5.38

Find the quotient: (18x3y−36xy2)÷(−3xy).(18x3y−36xy2)÷(−3xy).

Solution

(18x3y−36xy2)÷(−3xy)(18x3y−36xy2)÷(−3xy)
Rewrite as a fraction. 18x3y−36xy2−3xy18x3y−36xy2−3xy
Divide each term by the divisor. Be careful with the signs! 18x3y−3xy−36xy2−3xy18x3y−3xy−36xy2−3xy
Simplify. −6x2+12y−6x2+12y
This table demonstrates the step-by-step process of dividing a polynomial by a monomial, showing the transformation of the mathematical expression.

Try It 5.75

Find the quotient: (32a2b−16ab2)÷(−8ab).(32a2b−16ab2)÷(−8ab).

Try It 5.76

Find the quotient: (−48a8b4−36a6b5)÷(−6a3b3).(−48a8b4−36a6b5)÷(−6a3b3).

Divide Polynomials Using Long Division

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, 875, by a 2-digit number, 25.

This figure shows the long division of 875 divided by 25. 875 is labeled dividend and 25 is labeled divisor. The result of 35 is labeled quotient. The 3 in 35 is determined from the number of times we can divide 25 into 87. Multiplying 25 and 3 results in 75. 75 is subtracted from 87 to get 12. The 5 from 875 is dropped down to make 12 into 125. The 5 in 35 is determined from the number of times was can divide 25 into 125. Since 25 goes into 125 evenly there is no remainder. The result of subtracting 125 from 125 is 0 which is labeled remainder.

We check division by multiplying the quotient by the divisor.

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

35·25875✓35·25875✓

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 5.39

Find the quotient: (x2+9x+20)÷(x+5).(x2+9x+20)÷(x+5).

Solution

A mathematical expression showing polynomial division: (x^2 + 9x + 20) ÷ (x + 5).
Write it as a long division problem.
Be sure the dividend is in standard form.
A mathematical long division problem showing the polynomial x^2 + 9x + 20 being divided by the binomial x + 5.
Divide x2x2 by x.x. It may help to ask yourself, “What do I need
to multiply xx by to get x2x2?”
Polynomial long division problem where x^2 + 9x + 20 is being divided by x + 5, with 'x' as the first term of the quotient.
Put the answer, x,x, in the quotient over the xx term.
Multiply xx times x+5.x+5. Line up the like terms under the dividend.
A step in polynomial long division, showing (x+5) dividing into x^2 + 9x + 20. The term 'x' is placed in the quotient, and x(x+5) = x^2 + 5x is written below, ready for subtraction.
Subtract x2+5xx2+5x from x2+9x.x2+9x.
You may find it easier to change the signs and then add.
Then bring down the last term, 20.
Polynomial long division: dividing x^2 + 9x + 20 by x + 5. After placing 'x' in the quotient, -x^2 + (-5x) is subtracted from the dividend, resulting in 4x + 20, highlighted in red.

Divide 4x4x by x.x. It may help to ask yourself, “What do I
need to multiply xx by to get 4x4x?”
Put the answer, 44, in the quotient over the constant term.
Polynomial long division of (x^2 + 9x + 20) by (x + 5) is demonstrated, showing the initial steps to arrive at the quotient x + 4. The calculation subtracts (-x^2 + (-5x)) to get 4x + 20.
Multiply 4 times x+5.x+5. Polynomial long division of x^2 + 9x + 20 by x + 5, yielding a quotient of x + 4. The steps illustrate the process of dividing polynomials, from the initial division to the final remainder.
Subtract 4x+204x+20 from 4x+20.4x+20. Polynomial long division problem for (x^2 + 9x + 20) divided by (x + 5), with the steps leading to a quotient of (x + 4) and a remainder of 0.
Check:
Multiply the quotient by the divisor. (x+4)(x+5)(x+4)(x+5)
You should get the dividend. x2+9x+20✓x2+9x+20✓

Try It 5.77

Find the quotient: (y2+10y+21)÷(y+3).(y2+10y+21)÷(y+3).

Try It 5.78

Find the quotient: (m2+9m+20)÷(m+4).(m2+9m+20)÷(m+4).

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

Look back at the dividends in previous examples. The terms were written in descending order of degrees, and there were no missing degrees. The dividend in this example will be x4−x2+5x−6.x4−x2+5x−6. It is missing an x3x3 term. We will add in 0x30x3 as a placeholder.

Example 5.40

Find the quotient: (x4−x2+5x−6)÷(x+2).(x4−x2+5x−6)÷(x+2).

Solution

Notice that there is no x3x3 term in the dividend. We will add 0x30x3 as a placeholder.

A mathematical expression showing the division of a polynomial (x^4 - x^2 + 5x - 6) by a binomial (x + 2).
Write it as a long division problem. Be sure the dividend is in standard form with placeholders for missing terms. A polynomial long division setup, showing x^4 + 0x^3 - x^2 + 5x - 6 being divided by x + 2, commonly used to find quotients and remainders of polynomial expressions.
Divide x4x4 by x.x.
Put the answer, x3,x3, in the quotient over the x3x3 term.
Multiply x3x3 times x+2.x+2. Line up the like terms.
Subtract and then bring down the next term.
A step in polynomial long division where x^4 + 0x^3 - x^2 + 5x - 6 is divided by x + 2, showing the subtraction of (x^4 + 2x^3) from x^4 + 0x^3 to get -2x^3 - x^2. A note suggests changing signs and adding.
Divide −2x3−2x3 by x.x.
Put the answer, −2x2,−2x2, in the quotient over the x2x2 term.
Multiply −2x2−2x2 times x+1.x+1. Line up the like terms
Subtract and bring down the next term.
A visual guide to polynomial long division, showing the steps to divide x^4 - x^2 + 5x - 6 by x + 2. It highlights the subtraction of terms, noting that changing signs and adding can be helpful.
Divide 3x23x2 by x.x.
Put the answer, 3x,3x, in the quotient over the xx term.
Multiply 3x3x times x+1.x+1. Line up the like terms.
Subtract and bring down the next term.
This image illustrates polynomial long division, providing a step-by-step example. A helpful tip suggests changing the signs and adding instead of directly subtracting terms during the process.
Divide −x−x by x.x.
Put the answer, −1,−1, in the quotient over the constant term.
Multiply −1−1 times x+1.x+1. Line up the like terms.
Change the signs, add.

Write the remainder as a fraction with the divisor as the denominator.
This image illustrates the process of polynomial long division, dividing x^4 - x^2 + 5x - 6 by x + 2, with a helpful reminder to change signs when subtracting.
To check, multiply (x+2)(x3−2x2+3x−1−4x+2)(x+2)(x3−2x2+3x−1−4x+2).
The result should be x4−x2+5x−6.x4−x2+5x−6.

Try It 5.79

Find the quotient: (x4−7x2+7x+6)÷(x+3).(x4−7x2+7x+6)÷(x+3).

Try It 5.80

Find the quotient: (x4−11x2−7x−6)÷(x+3).(x4−11x2−7x−6)÷(x+3).

In the next example, we will divide by 2a+3.2a+3. As we divide, we will have to consider the constants as well as the variables.

Example 5.41

Find the quotient: (8a3+27)÷(2a+3).(8a3+27)÷(2a+3).

Solution

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

A mathematical expression showing the division of a sum of cubes, (8a^3 + 27), by a binomial, (2a + 3), written horizontally with a division symbol.
A step-by-step example of polynomial long division showing (8a^3 + 27) divided by (2a + 3), resulting in 4a^2 - 6a + 9 with a remainder of 0. The intermediate multiplication steps are also indicated.

To check, multiply (2a+3)(4a2−6a+9).(2a+3)(4a2−6a+9).

The result should be 8a3+27.8a3+27.

Try It 5.81

Find the quotient: (x3−64)÷(x−4).(x3−64)÷(x−4).

Try It 5.82

Find the quotient: (125x3−8)÷(5x−2).(125x3−8)÷(5x−2).

Divide Polynomials using Synthetic Division

As we have mentioned before, mathematicians like to find patterns to make their work easier. Since long division can be tedious, let’s look back at the long division we did in Example 5.39 and look for some patterns. We will use this as a basis for what is called synthetic division. The same problem in the synthetic division format is shown next.

The figure shows the long division of 1 x squared plus 9 x plus 20 divided by x plus 5 right next to the same problem done with synthetic division. In the long division problem, the coefficients of the dividend are 1 and 9 and 20 and the zero of the divisor is negative 5. In the synthetic division problem, we just write the numbers negative 5 1 9 20 with a line separating the negative 5. In the long division problem, the subtracted terms are 5 x and 20. In the synthetic division problem the second line is the numbers negative 5 and negative 20. The remainder of the problem is 0 and the quotient is x plus 4. The synthetic division puts these coefficients as the last line 1 4 0.

Synthetic division basically just removes unnecessary repeated variables and numbers. Here all the xx and x2x2 are removed. as well as the −x2−x2 and −4x−4x as they are opposite the term above.

The first row of the synthetic division is the coefficients of the dividend. The −5−5 is the opposite of the 5 in the divisor.

The second row of the synthetic division are the numbers shown in red in the division problem.

The third row of the synthetic division are the numbers shown in blue in the division problem.

Notice the quotient and remainder are shown in the third row.

Synthetic division only works when the divisor is of the formx−c.Synthetic division only works when the divisor is of the formx−c.

The following example will explain the process.

Example 5.42

Use synthetic division to find the quotient and remainder when 2x3+3x2+x+82x3+3x2+x+8 is divided by x+2.x+2.

Solution

Write the dividend with decreasing powers of x.x. The polynomial expression 2x^3 + 3x^2 + x + 8.
Write the coefficients of the terms as the first
row of the synthetic division.
A sequence of numbers 2, 3, 1, 8 is displayed on a white background, partially enclosed by a dark gray bracket on the left.
Write the divisor as x−cx−c and place c
in the synthetic division in the divisor box.
A synthetic division problem showing -2 as the divisor and 2, 3, 1, 8 as the coefficients of a polynomial.
Bring down the first coefficient to the third row. Illustrates the initial step of synthetic division: the first coefficient (2) of the polynomial is brought down below the line, with -2 as the divisor and 3, 1, 8 as subsequent coefficients.
Multiply that coefficient by the divisor and place the
result in the second row under the second coefficient.
An image showing the initial steps of synthetic division. The number -2 is being divided into a polynomial with coefficients 2, 3, 1, and 8. The leading coefficient, 2, is brought down, then multiplied by -2 to get -4.
Add the second column, putting the result in the third row. A mathematical division problem is depicted, likely synthetic division, showing the numbers -2, 2, 3, 1, 8. Below the 3, -4 is written, and below a horizontal line, -1 appears with a light blue arrow pointing down from 3 to -4 and then to -1, indicating subtraction. The number 2 is also shown below the first horizontal line.
Multiply that result by the divisor and place the
result in the second row under the third coefficient.
An image illustrating the initial steps of synthetic division, showing coefficients (2, 3, 1, 8) divided by -2, with multiplication and addition leading to the first few quotient terms.
Add the third column, putting the result in the third row. A step in synthetic division showing the coefficients 2, 3, 1, and 8 being divided by -2. The process shows bringing down 2, multiplying by -2 to get -4, then adding 3 and -4 to get -1. Next, -1 is multiplied by -2 to get 2, and 1 plus 2 yields 3.
Multiply that result by the divisor and place the
result in the third row under the third coefficient.
A mathematical example demonstrating the process of synthetic division, with a divisor of -2 and polynomial coefficients 2, 3, 1, and 8. Blue arrows indicate the steps of multiplication and addition.
Add the final column, putting the result in the third row. An example of synthetic division is shown, with -2 as the divisor and 2, 3, 1, 8 as the dividend coefficients. The result is a quotient with coefficients 2, -1, 3 and a remainder of 2.
The quotient is 2x2−1x+32x2−1x+3 and the remainder is 2.

The division is complete. The numbers in the third row give us the result. The 2−132−13 are the coefficients of the quotient. The quotient is 2x2−1x+3.2x2−1x+3. The 2 in the box in the third row is the remainder.

Check:

(quotient)(divisor)+remainder=dividend(2x2−1x+3)(x+2)+2=?2x3+3x2+x+8 2x3−x2+3x+4x2−2x+6+2=?2x3+3x2+x+8 2x3+3x2+x+8=2x3+3x2+x+8✓ (quotient)(divisor)+remainder=dividend(2x2−1x+3)(x+2)+2=?2x3+3x2+x+8 2x3−x2+3x+4x2−2x+6+2=?2x3+3x2+x+8 2x3+3x2+x+8=2x3+3x2+x+8✓

Try It 5.83

Use synthetic division to find the quotient and remainder when 3x3+10x2+6x−23x3+10x2+6x−2 is divided by x+2.x+2.

Try It 5.84

Use synthetic division to find the quotient and remainder when 4x3+5x2−5x+34x3+5x2−5x+3 is divided by x+2.x+2.

In the next example, we will do all the steps together.

Example 5.43

Use synthetic division to find the quotient and remainder when x4−16x2+3x+12x4−16x2+3x+12 is divided by x+4.x+4.

Solution

The polynomial x4−16x2+3x+12x4−16x2+3x+12 has its term in order with descending degree but we notice there is no x3x3 term. We will add a 0 as a placeholder for the x3x3 term. In x−cx−c form, the divisor is x−(−4).x−(−4).

The figure shows the results of using synthetic division with the example of the polynomial x to the fourth power minus 16 x squared plus 3 x plus 12 divided by x plus 4. The divisor number if negative 4. The first row is 1 0 negative 16 3 12. The first column is 1 blank 1. The second column is negative 16 16 0. The third column is 3 0 3. The fourth column is 12 negative 12 0.

We divided a 4th degree polynomial by a 1st degree polynomial so the quotient will be a 3rd degree polynomial.

Reading from the third row, the quotient has the coefficients 1−403,1−403, which is x3−4x2+3.x3−4x2+3. The remainder
is 0.

Try It 5.85

Use synthetic division to find the quotient and remainder when x4−16x2+5x+20x4−16x2+5x+20 is divided by x+4.x+4.

Try It 5.86

Use synthetic division to find the quotient and remainder when x4−9x2+2x+6x4−9x2+2x+6 is divided by x+3.x+3.

Divide Polynomial Functions

Just as polynomials can be divided, polynomial functions can also be divided.

Division of Polynomial Functions

For functions f(x)f(x) and g(x),g(x), where g(x)≠0,g(x)≠0,

(fg)(x)=f(x)g(x)(fg)(x)=f(x)g(x)

Example 5.44

For functions f(x)=x2−5x−14f(x)=x2−5x−14 and g(x)=x+2,g(x)=x+2, find: ⓐ (fg)(x)(fg)(x) ⓑ (fg)(−4).(fg)(−4).

Solution

ⓐ

Equation shows f over g of x equals f of x divided by g of x. This is translated into a division problem showing x squared minus 5x minus 14 divided by x plus 2. The quotient is x minus 7.


Substitute for f(x)f(x) and g(x).g(x). (fg)(x)=x2−5x−14x+2(fg)(x)=x2−5x−14x+2
Divide the polynomials. (fg)(x)=x−7(fg)(x)=x−7
This table illustrates the step-by-step polynomial division of two functions, f(x) and g(x), showing the substitution and the final simplified quotient.



ⓑ In part ⓐ we found (fg)(x)(fg)(x) and now are asked to find (fg)(−4).(fg)(−4).

(fg)(x)=x−7(fg)(x)=x−7
To find (fg)(−4),(fg)(−4), substitute x=−4.x=−4. (fg)(−4)=−4−7(fg)(−4)=−4−7
(fg)(−4)=−11(fg)(−4)=−11
This table demonstrates the step-by-step evaluation of the function (f/g)(x) at x=-4.

Try It 5.87

For functions f(x)=x2−5x−24f(x)=x2−5x−24 and g(x)=x+3,g(x)=x+3, find ⓐ (fg)(x)(fg)(x) ⓑ (fg)(−3).(fg)(−3).

Try It 5.88

For functions f(x)=x2−5x−36f(x)=x2−5x−36 and g(x)=x+4,g(x)=x+4, find ⓐ (fg)(x)(fg)(x) ⓑ (fg)(−5).(fg)(−5).

Use the Remainder and Factor Theorem

Let’s look at the division problems we have just worked that ended up with a remainder. They are summarized in the chart below. If we take the dividend from each division problem and use it to define a function, we get the functions shown in the chart. When the divisor is written as x−c,x−c, the value of the function at c,f(c),c,f(c), is the same as the remainder from the division problem.

Dividend Divisor x−cx−c Remainder Function f(c)f(c)
x4−x2+5x−6x4−x2+5x−6 x−(−2)x−(−2) −4−4 f(x)=x4−x2+5x−6f(x)=x4−x2+5x−6 −4−4
3x3−2x2−10x+83x3−2x2−10x+8 x−2x−2 4 f(x)=3x3−2x2−10x+8f(x)=3x3−2x2−10x+8 4
x4−16x2+3x+15x4−16x2+3x+15 x−(−4)x−(−4) 3 f(x)=x4−16x2+3x+15f(x)=x4−16x2+3x+15 3
Table 5.1

To see this more generally, we realize we can check a division problem by multiplying the quotient times the divisor and add the remainder. In function notation we could say, to get the dividend f(x),f(x), we multiply the quotient, q(x)q(x) times the divisor, x−c,x−c, and add the remainder, r.

A mathematical equation illustrating the Remainder Theorem: f(x) = q(x)(x - c) + r, where f(x) is a polynomial, q(x) is the quotient, (x-c) is the divisor, and r is the remainder.
If we evaluate this at c,c, we get: A mathematical equation is displayed: f(c) = q(c)(c - c) + r. The variable 'c' is highlighted in red within f(c), q(c), and (c - c).
A mathematical equation showing f(c) = q(c)(0) + r, where the term q(c)(0) simplifies to zero, resulting in f(c) = r.
The mathematical equation f(c) = r is displayed on a white background.

This leads us to the Remainder Theorem.

Remainder Theorem

If the polynomial function f(x)f(x) is divided by x−c,x−c, then the remainder is f(c).f(c).

Example 5.45

Use the Remainder Theorem to find the remainder when f(x)=x3+3x+19f(x)=x3+3x+19 is divided by x+2.x+2.

Solution

To use the Remainder Theorem, we must use the divisor in the x−cx−c form. We can write the divisor x+2x+2 as x−(−2).x−(−2). So, our cc is −2.−2.

To find the remainder, we evaluate f(c)f(c) which is f(−2).f(−2).

The image shows the mathematical function f(x) = x³ + 3x + 19.
To evaluate f(−2),f(−2), substitute x=−2.x=−2. A mathematical equation is displayed: f(-2) = (-2)^3 + 3(-2) + 19, showing the substitution of -2 into a function.
Simplify. A mathematical equation is displayed against a white background, reading 'f(-2) = -8 - 6 + 19'.
A mathematical equation displays 'f(-2) = 5' in a clear, bold font on a white background, representing the function f evaluated at -2 equals 5.
The remainder is 5 when f(x)=x3+3x+19f(x)=x3+3x+19 is divided by x+2.x+2.
Check:
Use synthetic division to check.
A synthetic division problem with a divisor of -2. The dividend's coefficients are 1, 0, 3, 19. The result shows quotient coefficients 1, -2, 7, and a remainder of 5.
The remainder is 5.

Try It 5.89

Use the Remainder Theorem to find the remainder when f(x)=x3+4x+15f(x)=x3+4x+15 is divided by x+2.x+2.

Try It 5.90

Use the Remainder Theorem to find the remainder when f(x)=x3−7x+12f(x)=x3−7x+12 is divided by x+3.x+3.

When we divided 8a3+278a3+27 by 2a+32a+3 in Example 5.41 the result was 4a2−6a+9.4a2−6a+9. To check our work, we multiply 4a2−6a+94a2−6a+9 by 2a+32a+3 to get 8a3+278a3+27.

(4a2−6a+9)(2a+3)=8a3+27(4a2−6a+9)(2a+3)=8a3+27

Written this way, we can see that 4a2−6a+94a2−6a+9 and 2a+32a+3 are factors of 8a3+27.8a3+27. When we did the division, the remainder was zero.

Whenever a divisor, x−c,x−c, divides a polynomial function, f(x),f(x), and resulting in a remainder of zero, we say x−cx−c is a factor of f(x).f(x).

The reverse is also true. If x−cx−c is a factor of f(x)f(x) then x−cx−c will divide the polynomial function resulting in a remainder of zero.

We will state this in the Factor Theorem.

Factor Theorem

For any polynomial function f(x),f(x),

  • if x−cx−c is a factor of f(x),f(x), then f(c)=0f(c)=0
  • if f(c)=0,f(c)=0, then x−cx−c is a factor of f(x)f(x)

Example 5.46

Use the Remainder Theorem to determine if x−4x−4 is a factor of f(x)=x3−64.f(x)=x3−64.

Solution

The Factor Theorem tells us that x−4x−4 is a factor of f(x)=x3−64f(x)=x3−64 if f(4)=0.f(4)=0.

f(x)=x3−64f(x)=x3−64
To evaluate f(4)f(4) substitute x=4.x=4. f(4)=43−64f(4)=43−64
Simplify. f(4)=64−64f(4)=64−64
Subtract. f(4)=0f(4)=0
Step-by-step evaluation of the function f(x) = x^3 - 64 at x=4.

Since f(4)=0,f(4)=0, x−4x−4 is a factor of f(x)=x3−64.f(x)=x3−64.

Try It 5.91

Use the Factor Theorem to determine if x−5x−5 is a factor of f(x)=x3−125.f(x)=x3−125.

Try It 5.92

Use the Factor Theorem to determine if x−6x−6 is a factor of f(x)=x3−216.f(x)=x3−216.

Media

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

Section 5.4 Exercises

Practice Makes Perfect

Divide Monomials

In the following exercises, divide the monomials.

288.

15 r 4 s 9 ÷ ( 15 r 4 s 9 ) 15 r 4 s 9 ÷ ( 15 r 4 s 9 )

289.

20 m 8 n 4 ÷ ( 30 m 5 n 9 ) 20 m 8 n 4 ÷ ( 30 m 5 n 9 )

290.

18 a 4 b 8 −27 a 9 b 5 18 a 4 b 8 −27 a 9 b 5

291.

45 x 5 y 9 −60 x 8 y 6 45 x 5 y 9 −60 x 8 y 6

292.

( 10 m 5 n 4 ) ( 5 m 3 n 6 ) 25 m 7 n 5 ( 10 m 5 n 4 ) ( 5 m 3 n 6 ) 25 m 7 n 5

293.

( −18 p 4 q 7 ) ( −6 p 3 q 8 ) −36 p 12 q 10 ( −18 p 4 q 7 ) ( −6 p 3 q 8 ) −36 p 12 q 10

294.

( 6 a 4 b 3 ) ( 4 a b 5 ) ( 12 a 2 b ) ( a 3 b ) ( 6 a 4 b 3 ) ( 4 a b 5 ) ( 12 a 2 b ) ( a 3 b )

295.

( 4 u 2 v 5 ) ( 15 u 3 v ) ( 12 u 3 v ) ( u 4 v ) ( 4 u 2 v 5 ) ( 15 u 3 v ) ( 12 u 3 v ) ( u 4 v )

Divide a Polynomial by a Monomial

In the following exercises, divide each polynomial by the monomial.

296.

( 9 n 4 + 6 n 3 ) ÷ 3 n ( 9 n 4 + 6 n 3 ) ÷ 3 n

297.

( 8 x 3 + 6 x 2 ) ÷ 2 x ( 8 x 3 + 6 x 2 ) ÷ 2 x

298.

( 63 m 4 − 42 m 3 ) ÷ ( −7 m 2 ) ( 63 m 4 − 42 m 3 ) ÷ ( −7 m 2 )

299.

( 48 y 4 − 24 y 3 ) ÷ ( −8 y 2 ) ( 48 y 4 − 24 y 3 ) ÷ ( −8 y 2 )

300.

66 x 3 y 2 − 110 x 2 y 3 − 44 x 4 y 3 11 x 2 y 2 66 x 3 y 2 − 110 x 2 y 3 − 44 x 4 y 3 11 x 2 y 2

301.

72 r 5 s 2 + 132 r 4 s 3 − 96 r 3 s 5 12 r 2 s 2 72 r 5 s 2 + 132 r 4 s 3 − 96 r 3 s 5 12 r 2 s 2

302.

10 x 2 + 5 x − 4 −5 x 10 x 2 + 5 x − 4 −5 x

303.

20 y 2 + 12 y − 1 −4 y 20 y 2 + 12 y − 1 −4 y

Divide Polynomials using Long Division

In the following exercises, divide each polynomial by the binomial.

304.

( y 2 + 7 y + 12 ) ÷ ( y + 3 ) ( y 2 + 7 y + 12 ) ÷ ( y + 3 )

305.

( a 2 − 2 a − 35 ) ÷ ( a + 5 ) ( a 2 − 2 a − 35 ) ÷ ( a + 5 )

306.

( 6 m 2 − 19 m − 20 ) ÷ ( m − 4 ) ( 6 m 2 − 19 m − 20 ) ÷ ( m − 4 )

307.

( 4 x 2 − 17 x − 15 ) ÷ ( x − 5 ) ( 4 x 2 − 17 x − 15 ) ÷ ( x − 5 )

308.

( q 2 + 2 q + 20 ) ÷ ( q + 6 ) ( q 2 + 2 q + 20 ) ÷ ( q + 6 )

309.

( p 2 + 11 p + 16 ) ÷ ( p + 8 ) ( p 2 + 11 p + 16 ) ÷ ( p + 8 )

310.

( 3 b 3 + b 2 + 4 ) ÷ ( b + 1 ) ( 3 b 3 + b 2 + 4 ) ÷ ( b + 1 )

311.

( 2 n 3 − 10 n + 28 ) ÷ ( n + 3 ) ( 2 n 3 − 10 n + 28 ) ÷ ( n + 3 )

312.

( z 3 + 1 ) ÷ ( z + 1 ) ( z 3 + 1 ) ÷ ( z + 1 )

313.

( m 3 + 1000 ) ÷ ( m + 10 ) ( m 3 + 1000 ) ÷ ( m + 10 )

314.

( 64 x 3 − 27 ) ÷ ( 4 x − 3 ) ( 64 x 3 − 27 ) ÷ ( 4 x − 3 )

315.

( 125 y 3 − 64 ) ÷ ( 5 y − 4 ) ( 125 y 3 − 64 ) ÷ ( 5 y − 4 )

Divide Polynomials using Synthetic Division

In the following exercises, use synthetic Division to find the quotient and remainder.

316.

x3−6x2+5x+14x3−6x2+5x+14 is divided by x+1x+1

317.

x3−3x2−4x+12x3−3x2−4x+12 is divided by x+2x+2

318.

2x3−11x2+11x+122x3−11x2+11x+12 is divided by x−3x−3

319.

2x3−11x2+16x−122x3−11x2+16x−12 is divided by x−4x−4

320.

x4−5x2+13x+3x4−5x2+13x+3 is divided by x+3x+3

321.

x4+x2+6x−10x4+x2+6x−10 is divided by x+2x+2

322.

2x4−9x3+5x2−3x−62x4−9x3+5x2−3x−6 is divided by x−4x−4

323.

3x4−11x3+2x2+10x+63x4−11x3+2x2+10x+6 is divided by x−3x−3

Divide Polynomial Functions

In the following exercises, divide.

324.

For functions f(x)=x2−13x+36f(x)=x2−13x+36 and g(x)=x−4,g(x)=x−4, find ⓐ (fg)(x)(fg)(x) ⓑ (fg)(−1)(fg)(−1)

325.

For functions f(x)=x2−15x+54f(x)=x2−15x+54 and g(x)=x−9,g(x)=x−9, find ⓐ (fg)(x)(fg)(x) ⓑ (fg)(−5)(fg)(−5)

326.

For functions f(x)=x3+x2−7x+2f(x)=x3+x2−7x+2 and g(x)=x−2,g(x)=x−2, find ⓐ (fg)(x)(fg)(x) ⓑ (fg)(2)(fg)(2)

327.

For functions f(x)=x3+2x2−19x+12f(x)=x3+2x2−19x+12 and g(x)=x−3,g(x)=x−3, find ⓐ (fg)(x)(fg)(x) ⓑ (fg)(0)(fg)(0)

328.

For functions f(x)=x2−3x+2f(x)=x2−3x+2 and g(x)=x+3,g(x)=x+3, find ⓐ (fg)(x)(fg)(x) ⓑ (fg)(3)(fg)(3)

329.

For functions f(x)=x2+2x−3f(x)=x2+2x−3 and g(x)=x+3g(x)=x+3 find ⓐ (fg)(x)(fg)(x)ⓑ (fg)(3)(fg)(3)

Use the Remainder and Factor Theorem

In the following exercises, use the Remainder Theorem to find the remainder.

330.

f(x)=x3−8x+7f(x)=x3−8x+7 is divided by x+3x+3

331.

f(x)=x3−4x−9f(x)=x3−4x−9 is divided by x+2x+2

332.

f(x)=2x3−6x−24f(x)=2x3−6x−24 divided by x−3x−3

333.

f(x)=7x2−5x−8f(x)=7x2−5x−8 divided by x−1x−1

In the following exercises, use the Factor Theorem to determine if x−cx−c is a factor of the polynomial function.

334.

Determine whether x+3x+3 a factor of x3+8x2+21x+18x3+8x2+21x+18

335.

Determine whether x+4x+4 a factor of x3+x2−14x+8x3+x2−14x+8

336.

Determine whether x−2x−2 a factor of x3−7x2+7x−6x3−7x2+7x−6

337.

Determine whether x−3x−3 a factor of x3−7x2+11x+3x3−7x2+11x+3

Writing Exercises

338.

James divides 48y+648y+6 by 6 this way: 48y+66=48y.48y+66=48y. What is wrong with his reasoning?

339.

Divide 10x2+x−122x10x2+x−122x and explain with words how you get each term of the quotient.

340.

Explain when you can use synthetic division.

341.

In your own words, write the steps for synthetic division for x2+5x+6x2+5x+6 divided by x−2.x−2.

Self Check

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

The figure shows a table with seven rows and four columns. The first row is a header row and it labels each column. The first column header is “I can…”, the second is "confidently", the third is “with some help”, “no minus I don’t get it!”. Under the first column are the phrases “divide monomials”, “divide a polynomial by using a monomial”, “divide polynomials using long division”, “divide polynomials using synthetic division”, “divide polynomial functions”, and “use the Remainder and Factor Theorem”. Under the second, third, fourth columns are blank spaces where the learner can check what level of mastery they have achieved.

ⓑ On a scale of 1-10, how would you rate your mastery of this section in light of your responses on the checklist? How can you improve this?