Dividing Polynomials

As has been seen, polynomials can be added, subtracted and multiplied. As it turns out, they can also be divided. For this, the methods of long division or synthetic division can be used.

Explanation

How to Interpret Long Division

In long division notation, the numerator is placed underneath the division symbol, and the denominator is placed to the left. The expression below shows the division $\frac{13}{2}.$ $\begin{array}{l}\begin{array}{r} 2 & \begin{array}{|l} \hline 13 \\ \end{array}\end{array}\end{array}$ When the division is complete, there is a number above the division sign. $\begin{array}{ r l} & \phantom{1}6\\[-0.4em] 2& \begin{array}{|l}\hline \phantom{1}13 \\ \end{array} \\ & -12 \\ \hline &\phantom{21.}{\color{#FF0000}{1}} \end{array}$ Sometimes, two values will not divide evenly. As a result, the quotient contains a remainder. Here, there is a remainder of $1,$ which is expressed as $1$ divided by $2,$ or $\frac{1}{2}.$ The quotient can be written as $\dfrac{13}{2}=6+\dfrac{{\color{#FF0000}{1}}}{2}.$

Concept

Terminology for Polynomial Long Division

The concepts of numerator, denominator, quotient, and remainder are also used when working with polynomial long division. If, for example, $x^2 + 4$ is divided by $x-1,$ the division is written as follows. $\begin{array}{l}\begin{array}{r} x-1 & \begin{array}{|l} \hline x^2+4 \\ \end{array}\end{array}\end{array}$ When the division is finished, the quotient is $x+1$ and the remainder is $3$ . Thus, the result is written as

(x+1)+\dfrac{3}{x-1}.

Method

Polynomial Long Division

To divide expressions where both the numerator and denominator are polynomials, polynomial long division can be used. The process is similar to when dividing real numbers. Consider the following division. $\dfrac{x^3+7x-2-3x^2}{x^2+x+11}$

1

Sort the terms and write as long division

To begin, sort the terms in the polynomial by their degrees, in descending order, and write the division. $\begin{array}{l}\begin{array}{r} x^2+x+11 & \begin{array}{|l} \hline x^3-3x^2+7x-2 \\ \end{array}\end{array}\end{array}$

2

Divide the first term in the numerator by the first term in the denominator

Divide the first term in the numerator by the first term in the denominator. In this case,

x^3

should be divided by

x^2.

The result,

x,

is written over the numerator.

\begin{array}{l}\begin{array}{r} {\color{#0000FF}{x^2}}+x+11 & \begin{array}{|l} \hline {\color{#0000FF}{x^3}}-3x^2+7x-2 \\ \end{array}\end{array}\end{array}

\dfrac{{\color{#0000FF}{x^3}}}{{\color{#0000FF}{x^2}}}= x

\begin{array}{r} {\color{#FF0000}{x}}\phantom{-3x^2+7x-2+.} \\ \begin{array}{r} {\color{#FF0000}{x^2+x+11}} & \begin{array}{|l}\hline x^3-3x^2+7x-2 \\ \end{array}\end{array}\end{array}

3

Multiply the quotient in Step

2

by the denominator

The quotient from the previous step is now multiplied by the denominator. This gives ${\color{#FF0000}{x}}\left({\color{#FF0000}{x^2+x+11}}\right)=x^3+x^2+11x.$

4

Subtract the product in Step

3

from the numerator

The product from the previous step is subtracted from the numerator.

\begin{array}{r} {\color{#FF0000}{x}}\phantom{-3x^2+7x-2+} \\ \begin{array}{r} {\color{#FF0000}{x^2+x+11}} & \begin{array}{|l}\hline x^3-3x^2+7x-2 \\ \end{array}\end{array}\end{array}

Subtract

{\color{#FF0000}{x}} \left({\color{#FF0000}{x^2+x+11}}\right) = x^3+x^2+11x

\begin{array}{r}x\phantom{-3x^2+7x-2+33} \\ \begin{array}{rl} x^2+x+11 & \begin{array}{|l}\hline \phantom{-(}x^3-3x^2+7x\phantom{1}-2 \\ \end{array}\\ & -\left(x^3+x^2+11x\right) \end{array}\end{array}

RemoveParSignsRemove parentheses and change signs

\begin{array}{r}x\phantom{-3x^2+7x-2+3} \\ \begin{array}{rl} x^2+x+11 & \begin{array}{|l}\hline \phantom{-}x^3-3x^2+7x-2 \\ \end{array}\\ & -x^3\phantom{3}-x^2-11x \end{array}\end{array}

SimpTermsSimplify terms

\begin{array}{r} x\phantom{+7x-13888} \\ \begin{array}{r} x^2+x+11 & \begin{array}{|l}\hline \text{-}4x^2-4x-2 \\ \end{array}\end{array}\end{array}

The simplification in the last step is clarified below.

Simplify

The expression calculated from the long division can now be interpreted as $x+\dfrac{\text{-}4x^2-4x-2}{x^2+x+11}.$ Since the numerator in the remaining fraction has the same degree as the denominator, perform Steps $2$ - $4$ again.

5

Repeat Steps to 2

-4

until the numerator has a lower degree than the denominator

Steps

2-4

are repeated until the polynomial in the numerator has a lower degree than the polynomial in the denominator.

\begin{array}{r} x\phantom{+7x-13888} \\ \begin{array}{r} {\color{#0000FF}{x^2}}+x+11 & \begin{array}{|l}\hline {\color{#0000FF}{\text{-}4x^2}}-4x-2 \\ \end{array}\end{array}\end{array}

\dfrac{{\color{#0000FF}{{\color{#0000FF}{\text{-} 4x^2}}}}}{{\color{#0000FF}{{\color{#0000FF}{x^2}}}}}= \text{-} 4

\begin{array}{r} x {\color{#FF0000}{\, - \, 4 }} \phantom{-1138888} \\ \begin{array}{r} {\color{#FF0000}{x^2+x+11}} & \begin{array}{|l}\hline \text{-}4x^2-4x-2 \\ \end{array}\end{array}\end{array}

Subtract

{\color{#FF0000}{\text{-} 4}} \left({\color{#FF0000}{x^2+x+11}}\right) = \text{-} 4x^2-4x-44

\begin{array}{r}x-4 \phantom{-(\text{-} 4x^2+12} \\ \begin{array}{rl} x^2+x+11 & \begin{array}{|l}\hline \phantom{-(}\text{-} 4x^2-4x-2 \\ \end{array}\\ & -\big(\text{-} 4x^2-4x-44\big) \end{array}\end{array}

RemoveParSignsRemove parentheses and change signs

\begin{array}{r}x-4 \phantom{\text{-} 4x^2+12} \\ \begin{array}{rl} x^2+x+11 & \begin{array}{|l}\hline \text{-} 4x^2-4x-2 \\ \end{array}\\ & +4x^2+4x+44 \end{array}\end{array}

SimpTermsSimplify terms

\begin{array}{r} x-4 \phantom{11} \\ \begin{array}{r} x^2+x+11 & \begin{array}{|l}\hline 42\phantom{4575} \\ \end{array}\end{array}\end{array}

Now, the numerator has a degree of

0

and the denominator has a degree of

2.

Thus, the degree of the numerator is lower than that of the denominator, meaning that the division is complete. When long division is performed on paper, it looks something like this.

The quotient of $\dfrac{x^3+7x-2-3x^2}{x^2+x+11}$ is the sum the expression above the division symbol and the remainder divided by the denominator. $x-4+\dfrac{42}{x^2+x+11}$

Method

Synthetic Division

When dividing polynomials by binomials of the form

x-k,

where

x

is the independent variable and

k

is a real number, there is an alternative to polynomial long division. Synthetic division uses

k

and the coefficients of the numerator to find the quotient.

\dfrac{\text{-} x^3+4x^2+9}{x-3}

The result of the example division above can be determined using this process.

1

Set up the division using

k

and the coefficients of the numerator

The division symbol for synthetic division is L-shaped. The number $k$ is written to the left. Here it's $3.$ $\begin{array}{l} \begin{array}{r} 3 & \begin{array}{|cc} & & & \\ \\ \hline \end{array} \end{array} \end{array}$ Next, write the coefficients of the numerator to the right. The numerator is $\text{-} x^3+4x^2+9$ so the coefficients are $\text{-}1,$ $4,$ $0,$ and $9.$ Note there is no $x$ -term so that coefficient is $0.$ $\begin{array}{l} \begin{array}{r} 3 & \begin{array}{|cc} \text{-}1&4 &0 &9 \\ \\ \hline \end{array} \end{array} \end{array}$

2

Bring down the first coefficient

First, bring down the left-most coefficient. Here, that is

\text{-}1,

so

\text{-}1

is written below the horizontal line.

\begin{array}{l} \begin{array}{r} 3 & \begin{array}{|cc} {\color{#FF0000}{\text{-}1}}&4 &0 &9 \\ \\ \hline \end{array} \end{array} \end{array}

Bring down

{\color{#FF0000}{\text{-}1}}

\begin{array}{l} \begin{array}{r} {\color{#FF0000}{3}} & \begin{array}{|cc} \text{-}1&4 &0 &9 \\ & & & \\ \hline \end{array}\\ & \begin{array}{c} {\color{#FF0000}{\text{-}1}} &\phantom{4} &\phantom{0} &\phantom{9} \end{array} \end{array} \end{array}

3

Multiply

k

and the new number below the line

The newest number under the line is now multiplied by

k.

Here that means

3(\text{-}1)=\text{-}3.

The result is placed under the second term in the numerator.

\begin{array}{l} \begin{array}{r} {\color{#FF0000}{3}} & \begin{array}{|cc} \text{-}1&4 &0 &9 \\ & & & \\ \hline \end{array}\\ & \begin{array}{c} {\color{#FF0000}{\text{-}1}} &\phantom{4} &\phantom{0} &\phantom{9} \end{array} \end{array} \end{array}

{\color{#FF0000}{3}} ({\color{#FF0000}{\text{-}1}})=\text{-}3

\begin{array}{l} \begin{array}{r} 3 & \begin{array}{|cc} \text{-}1&{\color{#FF0000}{4}} &0 &9 \\ & {\color{#FF0000}{\text{-}3}} & & \\ \hline \end{array}\\ & \begin{array}{c} \text{-}1 &\phantom{4} &\phantom{0} &\phantom{9} \end{array} \end{array} \end{array}

4

Add the numbers in the column above the line

Now, the numbers in the column created by multiplying are added.

\begin{array}{l} \begin{array}{r} 3 & \begin{array}{|cc} \text{-}1&{\color{#FF0000}{4}} &0 &9 \\ & {\color{#FF0000}{\text{-}3}} & & \\ \hline \end{array}\\ & \begin{array}{c} \text{-}1 &\phantom{4} &\phantom{0} &\phantom{9} \end{array} \end{array} \end{array}

{\color{#FF0000}{4}}+({\color{#FF0000}{\text{-}3}})=1

\begin{array}{l} \begin{array}{r} {\color{#FF0000}{3}} & \begin{array}{|cc} \text{-}1&4 &0 &9 \\ & \text{-}3 & & \\ \hline \end{array}\\ & \begin{array}{c} \text{-}1 &{\color{#FF0000}{1}} &\phantom{0} &\phantom{9} \end{array} \end{array} \end{array}

5

Repeat Steps

3

-

4

through the last column

These steps are repeated through the last column.

\begin{array}{l} \begin{array}{r} {\color{#FF0000}{3}} & \begin{array}{|cc} \text{-}1&4 &0 &9 \\ & \text{-}3 & & \\ \hline \end{array}\\ & \begin{array}{c} \text{-}1 &{\color{#FF0000}{1}} &\phantom{0} &\phantom{9} \end{array} \end{array} \end{array}

{\color{#FF0000}{3}}\cdot{\color{#FF0000}{1}}=3

\begin{array}{l} \begin{array}{r} 3 & \begin{array}{|cc} \text{-}1&4 &{\color{#FF0000}{0}} &9 \\ & \text{-}3 &{\color{#FF0000}{3}} & \\ \hline \end{array}\\ & \begin{array}{c} \text{-}1 &1 &\phantom{0} &\phantom{9} \end{array} \end{array} \end{array}

{\color{#FF0000}{0}}+{\color{#FF0000}{3}}=3

\begin{array}{l} \begin{array}{r} {\color{#FF0000}{3}} & \begin{array}{|cc} \text{-}1&4 &0 &9 \\ & \text{-}3 &3 & \\ \hline \end{array}\\ & \begin{array}{c} \text{-}1 &1 &{\color{#FF0000}{3}} &\phantom{9} \end{array} \end{array} \end{array}

{\color{#FF0000}{3}}\cdot{\color{#FF0000}{3}}=9

\begin{array}{l} \begin{array}{r} 3 & \begin{array}{|cc} \text{-}1&4 &0 &{\color{#FF0000}{9}} \\ & \text{-}3 &3 &{\color{#FF0000}{9}} \\ \hline \end{array}\\ & \begin{array}{c} \text{-}1 &1 &3 &\phantom{9} \end{array} \end{array} \end{array}

{\color{#FF0000}{9}}+{\color{#FF0000}{9}}=18

\begin{array}{l} \begin{array}{r} 3 & \begin{array}{|cc} \text{-}1&4 &0 &9\phantom{1} \\ & \text{-}3 &3 &9\phantom{1} \\ \hline \end{array}\\ & \begin{array}{c} \text{-}1 &1 &3 &18 \end{array} \end{array} \end{array}

Now that the division is complete, the numbers below the horizontal line are the coefficients of the quotient.

6

Write the quotient

Keeping in mind that the quotient is one degree lower than the numerator, the quotient can be written using the results of the synthetic division above. $\text{-} x^2+x+3+\dfrac{18}{x-3}$ Since the last number below the horizontal line was not $0,$ there is a remainder. The remainder is written as the quotient of the remainder and the denominator.

Rule

Remainder Theorem

If a polynomial $p(x)$ is divided by a binomial in the form $x-a$ , where $a$ is a real number, the result will be the sum of another polynomial and a fraction where the numerator is the remainder, $r.$ $\begin{gathered} \frac{p(x)}{x-a} = q(x) + \frac{r}{x-a} \end{gathered}$ Then, according to the Remainder Theorem, the remainder can be determined by $\begin{gathered} r=p(a). \end{gathered}$ In other words, the remainder is given by the function value of $p(x)$ when $x=a.$

Rule

r=p(a)

To show why the remainder can be given by

r=p(a),

the division

\begin{gathered} \frac{p(x)}{x-a} = q(x) + \frac{r}{x-a} \end{gathered}

can be rewritten by multiplying both sides by the binomial

x-a.

\dfrac{p(x)}{x-a} = q(x) + \dfrac{r}{x-a}

MultEqn

\text{LHS} \cdot (x-a)=\text{RHS}\cdot (x-a)

p(x)=\left(q(x)+\dfrac{r}{x-a}\right)\cdot{(x-a)}

DistrDistribute

(x-a)

p(x)=q(x)(x-a)+r

By evaluating

p(x)

when

x=a

the expression can be simplified to

r=p(a).

p(x)=q(x)(x-a)+r

Substitute

x={\color{#0000FF}{a}}

p({\color{#0000FF}{a}})=q({\color{#0000FF}{a}})({\color{#0000FF}{a}}-a)+r

SubTermSubtract term

p(a)=q(a)\cdot0+r

MultiplyMultiply

p(a)=r

RearrangeEqnRearrange equation

r=p(a)

Therefore, the remainder is

r=p(a).

The Remainder Theorem can be used to calculate the value of a polynomial function

p(x)

for a number

a.

By dividing the polynomial with the binomial

x-a,

the reminder will be equal to the function value

p(a).

If the remainder is zero,

\begin{gathered} p(a)=0, \end{gathered}

the binomial

x-a

is a factor of the polynomial

p(x).

Evaluate the polynomial using synthetic division

fullscreen

Exercise

Use synthetic division to evaluate $f(4)$ if $f(x)=3x^4-5x^3+x^2-x-3.$

Show Solution

Solution

To evaluate the polynomial, we could substitute

x=4

into the rule and simplify. However, the Remainder Theorem tells us that

f(4)

is the remainder when dividing

f(x)

by

x-4.

Thus, by determining the remainder of

\dfrac{3x^4-5x^3+x^2-x-3}{x-4},

will give

f(4).

As instructed, we'll use synthetic division. The coefficients are

3,

\text{-}5,

1,

\text{-}1,

and

\text{-}3.

\begin{array}{l} \begin{array}{r} 4 & \begin{array}{|cc} {\color{#FF0000}{3}}&\text{-}5&1&\text{-}1&\text{-}3 \\ \\ \hline \end{array} \end{array} \end{array}

Bring down

{\color{#FF0000}{3}}

\begin{array}{l} \begin{array}{r} {\color{#FF0000}{4}} & \begin{array}{|cc} 3&\text{-}5&1&\text{-}1&\text{-}3 \\ & & & & \\ \hline \end{array}\\ & \begin{array}{c} {\color{#FF0000}{3}} &\phantom{\text{-}5}&\phantom{1} &\phantom{\text{-}1}&\phantom{\text{-}3} \end{array} \end{array} \end{array}

{\color{#FF0000}{4}}\cdot{\color{#FF0000}{3}}=12

\begin{array}{l} \begin{array}{r} 4 & \begin{array}{|cc} 3&{\color{#FF0000}{\text{-}5}}&1&\text{-}1&\text{-}3 \\ &{\color{#FF0000}{12}} & & & \\ \hline \end{array}\\ & \begin{array}{c} 3 &\phantom{\text{-}5}&\phantom{1} &\phantom{\text{-}1}&\phantom{\text{-}3} \end{array} \end{array} \end{array}

{\color{#FF0000}{\text{-}5}}+{\color{#FF0000}{12}}=7

\begin{array}{l} \begin{array}{r} {\color{#FF0000}{4}} & \begin{array}{|cc} 3&\text{-}5&1&\text{-}1&\text{-}3 \\ &12 & & & \\ \hline \end{array}\\ & \begin{array}{c} 3 &{\color{#FF0000}{7}}\phantom{3}&\phantom{1} &\phantom{\text{-}1}&\phantom{\text{-}3} \end{array} \end{array} \end{array}

{\color{#FF0000}{4}}\cdot{\color{#FF0000}{7}}=28

\begin{array}{l} \begin{array}{r} 4 & \begin{array}{|cc} 3&\text{-}5&{\color{#FF0000}{1}}&\text{-}1&\text{-}3 \\ &12 &{\color{#FF0000}{28}} & & \\ \hline \end{array}\\ & \begin{array}{c} 3 &7\phantom{5}&\phantom{11} &\phantom{\text{-}1}&\phantom{\text{-}3} \end{array} \end{array} \end{array}

{\color{#FF0000}{1}}+{\color{#FF0000}{28}}=29

\begin{array}{l} \begin{array}{r} {\color{#FF0000}{4}} & \begin{array}{|cc} 3&\text{-}5&1&\text{-}1&\text{-}3 \\ &12 &28 & & \\ \hline \end{array}\\ & \begin{array}{c} 3 &7\phantom{2}&{\color{#FF0000}{29}} &\phantom{\text{-}1}&\phantom{\text{-}3} \end{array} \end{array} \end{array}

{\color{#FF0000}{4}}\cdot{\color{#FF0000}{29}}=116

\begin{array}{l} \begin{array}{r} 4 & \begin{array}{|cc} 3&\text{-}5&1&{\color{#FF0000}{\text{-}1}}&\text{-}3 \\ &12 &28 &{\color{#FF0000}{116}} & \\ \hline \end{array}\\ & \begin{array}{c} 3 &7\phantom{1}&29 &\phantom{111}&\phantom{\text{-}3} \end{array} \end{array} \end{array}

{\color{#FF0000}{\text{-}1}}+{\color{#FF0000}{116}}=115

\begin{array}{l} \begin{array}{r} {\color{#FF0000}{4}} & \begin{array}{|cc} 3&\text{-}5&1&\text{-}1&\text{-}3 \\ &12 &28 &116 & \\ \hline \end{array}\\ & \begin{array}{c} 3 &7&29 &{\color{#FF0000}{115}}&\phantom{\text{-}3} \end{array} \end{array} \end{array}

{\color{#FF0000}{4}}\cdot{\color{#FF0000}{115}}=460

\begin{array}{l} \begin{array}{r} 4 & \begin{array}{|cc} 3&\text{-}5&1&\text{-}1&{\color{#FF0000}{\text{-}3}} \\ &12 &28 &116 &{\color{#FF0000}{460}} \\ \hline \end{array}\\ & \begin{array}{c} 3 &7\phantom{3}&29 &115&\phantom{133} \end{array} \end{array} \end{array}

{\color{#FF0000}{\text{-}3}}+{\color{#FF0000}{460}}=457

\begin{array}{l} \begin{array}{r} 4 & \begin{array}{|cc} 3&\text{-}5&1&\text{-}1&{\color{#FF0000}{\text{-}3}} \\ &12 &28 &116 &{\color{#FF0000}{460}} \\ \hline \end{array}\\ & \begin{array}{c} 3 &7&29 &115&457 \end{array} \end{array} \end{array}

The remainder of the division is

457,

which means that

f(4)=457.

1

2

3

4

5

1

2

3

4

5

6

Rule

Example

Evaluate the polynomial using synthetic division