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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 132.\frac{13}{2}. 213 \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. 1621131221.1 \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 1,1, which is expressed as 11 divided by 2,2, or 12.\frac{1}{2}. The quotient can be written as 132=6+12. \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, x2+4x^2 + 4 is divided by x1,x-1, the division is written as follows. x1x2+4 \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+1x+1 and the remainder is 33. Thus, the result is written as

(x+1)+3x1. (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. x3+7x23x2x2+x+11 \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. x2+x+11x33x2+7x2 \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, x3x^3 should be divided by x2.x^2. The result, x,x, is written over the numerator.
x2+x+11x33x2+7x2\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}
x3x2=x\dfrac{{\color{#0000FF}{x^3}}}{{\color{#0000FF}{x^2}}}= x
x3x2+7x2+.x2+x+11x33x2+7x2\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 22 by the denominator

The quotient from the previous step is now multiplied by the denominator. This gives x(x2+x+11)=x3+x2+11x. {\color{#FF0000}{x}}\left({\color{#FF0000}{x^2+x+11}}\right)=x^3+x^2+11x.

4

Subtract the product in Step 33 from the numerator
The product from the previous step is subtracted from the numerator.
x3x2+7x2+x2+x+11x33x2+7x2\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 x(x2+x+11)=x3+x2+11x{\color{#FF0000}{x}} \left({\color{#FF0000}{x^2+x+11}}\right) = x^3+x^2+11x
x3x2+7x2+33x2+x+11(x33x2+7x12(x3+x2+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}
x3x2+7x2+3x2+x+11x33x2+7x2x33x211x\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}
x+7x13888x2+x+11-4x24x2\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+-4x24x2x2+x+11. 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 22-44 again.

5

Repeat Steps to 24-4 until the numerator has a lower degree than the denominator
Steps 242-4 are repeated until the polynomial in the numerator has a lower degree than the polynomial in the denominator.
x+7x13888x2+x+11-4x24x2\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}
-4x2x2=-4\dfrac{{\color{#0000FF}{{\color{#0000FF}{\text{-} 4x^2}}}}}{{\color{#0000FF}{{\color{#0000FF}{x^2}}}}}= \text{-} 4
x41138888x2+x+11-4x24x2\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 -4(x2+x+11)=-4x24x44{\color{#FF0000}{\text{-} 4}} \left({\color{#FF0000}{x^2+x+11}}\right) = \text{-} 4x^2-4x-44
x4(-4x2+12x2+x+11(-4x24x2(-4x24x44)\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}
x4-4x2+12x2+x+11-4x24x2+4x2+4x+44\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}
x411x2+x+11424575\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 00 and the denominator has a degree of 2.2. 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.
Polynomial division.svg

The quotient of x3+7x23x2x2+x+11 \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. x4+42x2+x+11 x-4+\dfrac{42}{x^2+x+11}

Method

Synthetic Division

When dividing polynomials by binomials of the form xk,x-k, where xx is the independent variable and kk is a real number, there is an alternative to polynomial long division. Synthetic division uses kk and the coefficients of the numerator to find the quotient. -x3+4x2+9x3 \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 kk and the coefficients of the numerator

The division symbol for synthetic division is L-shaped. The number kk is written to the left. Here it's 3.3. 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 -x3+4x2+9\text{-} x^3+4x^2+9 so the coefficients are -1,\text{-}1, 4,4, 0,0, and 9.9. Note there is no xx-term so that coefficient is 0.0. 3-1409 \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 -1,\text{-}1, so -1\text{-}1 is written below the horizontal line.
3-1409\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 -1{\color{#FF0000}{\text{-}1}}
3-1409-1409\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 kk and the new number below the line
The newest number under the line is now multiplied by k.k. Here that means 3(-1)=-3.3(\text{-}1)=\text{-}3. The result is placed under the second term in the numerator.
3-1409-1409\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(-1)=-3{\color{#FF0000}{3}} ({\color{#FF0000}{\text{-}1}})=\text{-}3
3-1409-3-1409\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.
3-1409-3-1409\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+(-3)=1{\color{#FF0000}{4}}+({\color{#FF0000}{\text{-}3}})=1
3-1409-3-1109\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 33-44 through the last column
These steps are repeated through the last column.
3-1409-3-1109\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}
31=3{\color{#FF0000}{3}}\cdot{\color{#FF0000}{1}}=3
3-1409-33-1109\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}
0+3=3{\color{#FF0000}{0}}+{\color{#FF0000}{3}}=3
3-1409-33-1139\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}
33=9{\color{#FF0000}{3}}\cdot{\color{#FF0000}{3}}=9
3-1409-339-1139\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}
9+9=18{\color{#FF0000}{9}}+{\color{#FF0000}{9}}=18
3-14091-3391-11318\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. -x2+x+3+18x3 \text{-} x^2+x+3+\dfrac{18}{x-3} Since the last number below the horizontal line was not 0,0, 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)p(x) is divided by a binomial in the form xax-a, where aa is a real number, the result will be the sum of another polynomial and a fraction where the numerator is the remainder, r.r. p(x)xa=q(x)+rxa\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 r=p(a).\begin{gathered} r=p(a). \end{gathered} In other words, the remainder is given by the function value of p(x)p(x) when x=a.x=a.

Rule

info
r=p(a)r=p(a)
To show why the remainder can be given by r=p(a),r=p(a), the division p(x)xa=q(x)+rxa\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 xa.x-a.
p(x)xa=q(x)+rxa\dfrac{p(x)}{x-a} = q(x) + \dfrac{r}{x-a}
p(x)=(q(x)+rxa)(xa)p(x)=\left(q(x)+\dfrac{r}{x-a}\right)\cdot{(x-a)}
p(x)=q(x)(xa)+rp(x)=q(x)(x-a)+r
By evaluating p(x)p(x) when x=ax=a the expression can be simplified to r=p(a).r=p(a).
p(x)=q(x)(xa)+rp(x)=q(x)(x-a)+r
p(a)=q(a)(aa)+rp({\color{#0000FF}{a}})=q({\color{#0000FF}{a}})({\color{#0000FF}{a}}-a)+r
p(a)=q(a)0+rp(a)=q(a)\cdot0+r
p(a)=rp(a)=r
r=p(a)r=p(a)
Therefore, the remainder is r=p(a).r=p(a).
The Remainder Theorem can be used to calculate the value of a polynomial function p(x)p(x) for a number a.a. By dividing the polynomial with the binomial xa,x-a, the reminder will be equal to the function value p(a).p(a). If the remainder is zero, p(a)=0,\begin{gathered} p(a)=0, \end{gathered} the binomial xax-a is a factor of the polynomial p(x).p(x).
Exercise

Use synthetic division to evaluate f(4)f(4) if f(x)=3x45x3+x2x3. f(x)=3x^4-5x^3+x^2-x-3.

Solution
To evaluate the polynomial, we could substitute x=4x=4 into the rule and simplify. However, the Remainder Theorem tells us that f(4)f(4) is the remainder when dividing f(x)f(x) by x4.x-4. Thus, by determining the remainder of 3x45x3+x2x3x4, \dfrac{3x^4-5x^3+x^2-x-3}{x-4}, will give f(4).f(4). As instructed, we'll use synthetic division. The coefficients are 3,3, -5,\text{-}5, 1,1, -1,\text{-}1, and -3.\text{-}3.
43-51-1-3\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 3{\color{#FF0000}{3}}
43-51-1-33-51-1-3\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}
43=12{\color{#FF0000}{4}}\cdot{\color{#FF0000}{3}}=12
43-51-1-3123-51-1-3\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}
-5+12=7{\color{#FF0000}{\text{-}5}}+{\color{#FF0000}{12}}=7
43-51-1-3123731-1-3\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}
47=28{\color{#FF0000}{4}}\cdot{\color{#FF0000}{7}}=28
43-51-1-3122837511-1-3\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}
1+28=29{\color{#FF0000}{1}}+{\color{#FF0000}{28}}=29
43-51-1-3122837229-1-3\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}
429=116{\color{#FF0000}{4}}\cdot{\color{#FF0000}{29}}=116
43-51-1-3122811637129111-3\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}
-1+116=115{\color{#FF0000}{\text{-}1}}+{\color{#FF0000}{116}}=115
43-51-1-312281163729115-3\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}
4115=460{\color{#FF0000}{4}}\cdot{\color{#FF0000}{115}}=460
43-51-1-3122811646037329115133\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}
-3+460=457{\color{#FF0000}{\text{-}3}}+{\color{#FF0000}{460}}=457
43-51-1-312281164603729115457\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 457,457, which means that f(4)=457.f(4)=457.
info Show solution Show solution
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