6.6: Polynomial division
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Dividing polynomials is a process very similar to long division of whole numbers. Before we look at long division with polynomials, we will first master dividing a polynomial by a monomial.
Polynomial Division with Monomials
We divide a polynomial by a monomial by rewriting the expression as separated fractions rather than one fraction. We use the fact
\[\dfrac{a+b}{c}=\dfrac{a}{c}+\dfrac{b}{c}\nonumber\]
Divide: \(\dfrac{9x^5+6x^4-18x^3-24x^2}{3x^2}\)
Solution
Notice we have four terms with the same denominator. We can rewrite this expression as \(4\) fractions with the same denominator and then simplify.
\[\begin{array}{rl}\dfrac{9x^5+6x^4-18x^3-24x^2}{3x^2}&\text{Rewrite as four fractions with the same denominator} \\ \dfrac{9x^5}{3x^2}+\dfrac{6x^4}{3x^2}-\dfrac{18x^3}{3x^2}-\dfrac{24x^2}{3x^2}&\text{Reduce and apply the quotient rule of exponents} \\ 3x^3+2x^2-6x-8&\text{Quotient}\end{array}\nonumber\]
Divide: \(\dfrac{8x^3+4x^2-2x+6}{4x^2}\)
Solution
Notice we have four terms with the same denominator. We can rewrite this expression as \(4\) fractions with the same denominator and then simplify.
\[\begin{array}{rl}\dfrac{8x^3+4x^2-2x+6}{4x^2}&\text{Rewrite as four fractions with the same denominator} \\ \dfrac{8x^3}{4x^2}+\dfrac{4x^2}{4x^2}-\dfrac{2x}{4x^2}+\dfrac{6}{4x^2}&\text{Reduce and apply the quotient rule of exponents} \\ 2x+1-\dfrac{1}{2}x^{-1}+\dfrac{3}{2}x^{-2}&\text{Rewrite with positive exponents} \\ 2x+1-\dfrac{1}{2x}+\dfrac{3}{2x^2}&\text{Quotient}\end{array}\nonumber\]
Notice that sometimes we have fractions in the quotient. As long as the fractions are reduced, it is correct. Also, the second term \(\dfrac{4x^2}{4x^2}\) reduced completely to one.
Polynomial Division with Polynomials
Long division is required when we divide by a polynomial, i.e., when there is a sum or difference of terms in the denominator. Long division with polynomials works similar to long division with whole numbers. Let’s review an example.
Divide: \(631\div 4\)
Solution
Let’s review this example. The divisor is \(4\) and the dividend is \(631\). The answer is called the quotient. First we rewrite the division with the divisor on the outside, then the long division symbol, and the dividend inside the long division symbol.
\[\begin{aligned} \;\;1\;\; 5\;\; 7 \\ 4\overline{)\;\;6\;\; 3\;\; 1}&\qquad \text{How many times does }4\text{ divide into }6\text{?} \\ \underline{-4\qquad}&\qquad \text{Once. We write }1\text{ over the }6\text{, keeping place values.}\\ 2\;\;\;3\quad\; &\qquad\text{Bring down the next place value, }3\text{, and how many times does }4\text{ divide into }23\text{?} \\ \underline{-2\;\;0\quad}&\qquad 5\text{ times. We write }5\text{ over the }3\text{, keeping place values.} \\ 3\;\; 1 \\ \underline{-\;\;2\;\;8}&\qquad\text{Bring down the next place value, }1\text{, and how many times does }4\text{ divide into }31\text{?} \\ 3&\qquad 7\text{ times. We write }7\text{ over the }1\text{, keeping place values.} \end{aligned}\]
Hence, \(3\) is the remainder. So, we write the answer as the quotient, plus the remainder as a fraction:
\[157+\dfrac{3}{4}\nonumber\]
Simplifying this sum, we write \(157\dfrac{3}{4}\).
This method may seem elementary, but it isn’t the arithmetic we want to review, it is the method. We use the same method as we did in arithmetic, but now with polynomials.
When writing the answer with a remainder, we write the answer as \[\text{quotient}+\dfrac{\text{remainder}}{\text{divisor}}\nonumber\]
Recall, if we are given \(a\div b=c\), then \[\underset{\color{blue}{\text{dividend}}}{\underbrace{a}}\color{black}{}\div\underset{\color{blue}{\text{divisor}}}{\underbrace{b}}\color{black}{}=\underset{\color{blue}{\text{quotient}}}{\underbrace{c}}\nonumber\]
Divide: \(\dfrac{3x^3-5x^2-32x+7}{x-4}\).
Solution
Let’s start by writing the division as long division:
\[x\; -\; 4)\overline{\;\;3x^3\;\;-5x^2\;\;-32x\;\;+7}\nonumber\]
Now, we follow the same method as we did for arithmetic. Be sure to keep place values and change the signs for subtracting.
\[\begin{aligned}3x^2\;\;\;\; +7x\;\;-4 \\ x\; -\; 4\overline{)\quad\;\; 3x^3\;\;-5x^2\;\;-32x\;\;+7}&\qquad\text{How many times does }x\text{ divide into }3x^3\text{?} \\ \underline{-\;\; (3x^3\;-12x^2)\qquad\qquad\quad}&\qquad \text{Multiply }3x^2\text{ and }(x-4)\text{, then subtract.} \\ 7x^2\;\;-32x\;\qquad &\qquad\text{Bring down the next place value} \\ \underline{-(7x^2\;\;-28x)\qquad}&\qquad\text{How many times does }x\text{ divide into }7x^2\text{?}\:\color{blue}{7x} \\ -4x\;\;+7&\qquad\text{Bring down the next place value} \\ \underline{-(-4x)\;+16)}&\qquad\text{How many times does }x\text{ divide into }-4x\text{?}\:\color{blue}{-4} \\ -9&\qquad\text{Remainder} \end{aligned}\]
We can see now that the method just repeats itself until we obtain a value that the divisor doesn’t divide into and we obtain a remainder. Since the remainder is \(−9\), then we have
\[3x^2+7x-4-\dfrac{9}{x-4}\nonumber\]
Divide: \(\dfrac{6x^3-8x^2+10x+103}{2x+4}\)
Solution
Following the same pattern as before, we rewrite the division as long division and then complete the long division process. However, in this example, we will,“ Draw a line and change the signs,” so that we distribute the subtraction right away.
\[\begin{aligned}3x^2\;\;\; -10x\;\;\;+25 \\ 2x\; +\; 4\overline{)\quad\;\; 6x^3\;\;-8x^2\;\;+10x\;\;+103} \\ \underline{ -6x^3\;-12x^2\qquad\qquad\qquad\;} \\ -20x^2\;\;+10x\;\;\qquad \\ \underline{20x^2\;\;+40x\qquad\;\;} \\ 50x\;\;+103 \\ \underline{-50x\;-100} \\ 3 \end{aligned}\]
Since the remainder is \(3\), then we have
\[3x^2-10x+25+\dfrac{3}{2x+4}\nonumber\]
In Example 6.6.5 , instead of writing the subtraction sign with each step, we changed the terms to its opposite since that is essentially what we did in the previous example. We say,“ Draw a line, and change the signs,” for drawing the horizontal bar and the subtraction. This way, we can just add vertically. It is up to the discretion of the student which way to subtract, either subtract directly, or add the opposites
Polynomial Division with Missing Terms
Sometimes when dividing with polynomials, there may be a missing term in the dividend. We do not ignore the term, we just write in \(0\) as the coefficient.
Divide: \(\dfrac{2x^3-4x+42}{x+3}\)
Solution
We rewrite the division as long division and follow the same method, but put in zero for the coefficient. In this case, we are missing the \(x^2\) term; hence, we will put \(0x^2\) for that term and then divide as usual.
\[\begin{aligned}2x^2\;\;\; -6x\;\;\;+14 \\ x\; +\; 3\overline{)\;\; 2x^3\;\;\color{blue}{+0x^2}\;\;\;\color{black}{-4x}\;\;\:+42} \\ \underline{ -2x^3\;-6x^2\qquad\qquad\quad\;\;} \\ -6x^2\;\;-4x\;\;\qquad \\ \underline{6x^2\;\;+18x\qquad\;\;} \\ 14x\;\;+42 \\ \underline{-14x\;-42} \\ 0 \end{aligned}\]
Since the remainder is \(0\), then we have
\[2x^2-6x+14\nonumber\]
It is important to take a moment to check each problem to verify that the exponents descend and there are no missing terms. If so, we will have to adjust the problem as we did in Example 6.6.6 . Also, this final example illustrates, just as in regular long division, sometimes we have no remainder in a long division problem.
Polynomial Division with Functions
We can divide two polynomial functions the same way we divide polynomial expressions, except, now, we have functions. The method is the same, but the notation and problems change.
If \(f\) and \(g\) are two functions of \(x\), then \[(f\div g)(x)=\dfrac{f(x)}{g(x)}\nonumber\] where \(x\) is in the domain of \(f\) and \(g\), and \(g(x)\neq 0\).
Let \(f(x)=x^2-4x-5\) and \(g(x)=x-5\). Find \((f\div g)(x)\).
Solution
We start by applying the definition, then simplify completely.
\[\begin{aligned}(f\div g)(x)&=\dfrac{f(x)}{g(x)} \\ (f\div g)(x)&=\dfrac{x^2-4x-5}{x-5}\end{aligned}\]
Taking the divisor \(x − 5\), and the dividend to be \(x^2 − 4x − 5\), we get
\[\begin{aligned} x\;\;\;+1 \\ x\; -\; 5\overline{)\;\;x^2\;\;-4x\;\;-5} \\ \underline{-x^2\;\;+5x\qquad\;} \\ x\;\; -5 \\ \underline{-x\;\;+5} \\ 0 \end{aligned}\]
Since there is no remainder, then \((f\div g)(x) = x + 1\).
Polynomial Division by Synthetic Division
Another way to divide polynomials given a binomial divisor of the form \(x−a\) is using solely the coefficients of the polynomials. Synthetic division is a method for dividing polynomials without using the given variables of the polynomials, but only the coefficients.
Paolo Ruffini was an Italian mathematician of the early \(19^{\text{th}}\) century. In 1809, he was the first to describe a process called synthetic division, which could also be used to divide polynomials. It is also called division through Ruffini’s rule.
Let’s take Example 6.6.4 , and apply synthetic division to obtain the same results as we did with polynomial division. Divide: \(\dfrac{3x^3-5x^2-32x+7}{x-4}\)
Solution
First we take the excluded value of the expression, which is when \(x = 4\). We put this in the top right corner of the synthetic division table:
\[\begin{aligned} 4|& \\ &\: \underline{\qquad\qquad\qquad} \end{aligned}\]
Then, following the excluded value, we place the coefficients of the dividend in the same top row, in standard order:
\[\begin{aligned}4|&\;\;3\;\;-5\;\;-32\;\;7 \\ &\:\underline{\qquad\qquad\qquad\quad}\end{aligned}\]
Bring the leading coefficient down to the bottom row:
\[\begin{aligned}4|&\;\;3\;\;-5\;\;-32\;\;7 \\ &\:\underline{\downarrow \qquad\qquad\qquad} \\ &\;\;\color{blue}{3}\qquad\qquad\qquad \end{aligned}\]
Multiply the excluded value \(x = 4\) with the leading coefficient, i.e, \(4\cdot 3 = 12\), and put the product under the second coefficient:
\[\begin{array}{llll|l} 4|&3&-5&32&7 \\ &\downarrow&12 \\ \hline &\color{blue}{3}\end{array}\nonumber\]
Add down the coefficient and the product, i.e., add \(−5 + 12 = 7\), and place on the bottom row next to the leading coefficient:
\[\begin{array}{llll|l} 4|&3&-5&32&7 \\ &\downarrow&12 \\ \hline &\color{blue}{3}&\color{blue}{7}\end{array}\nonumber\]
Multiply the excluded value \(x = 4\) with the \(7\), i.e, \(4\cdot 7 = 28\), and put the product under the third coefficient:
\[\begin{array}{llll|l} 4|&3&-5&32&7 \\ &\downarrow&12&28 \\ \hline &\color{blue}{3}&\color{blue}{7}\end{array}\nonumber\]
Add down the coefficient and the product, i.e., add \(−32 + 28 = −4\), and place on the bottom row next to the \(7\):
\[\begin{array}{llll|l} 4|&3&-5&32&7 \\ &\downarrow&12&28 \\ \hline &\color{blue}{3}&\color{blue}{7}&\color{blue}{-4}\end{array}\nonumber\]
Multiply the excluded value \(x = 4\) with the \(−4\), i.e, \(4\cdot −4 = −16\), and put the product under the fourth coefficient:
\[\begin{array}{llll|l} 4|&3&-5&32&7 \\ &\downarrow&12&28&-16 \\ \hline &\color{blue}{3}&\color{blue}{7}&\color{blue}{-4}\end{array}\nonumber\]
Add down the coefficient and the product, i.e., add \(7 + (−16) = −9\), and place on the bottom row next to the \(4\):
\[\begin{array}{llll|l} 4|&3&-5&32&7 \\ &\downarrow&12&28&-16 \\ \hline &\color{blue}{3}&\color{blue}{7}&\color{blue}{-4}&\color{red}{-9}\end{array}\nonumber\]
The first three numbers in the last row of our synthetic division table are the coefficients of the quotient polynomial. Remember, we started with a third-degree polynomial and divided by a first-degree polynomial, so the quotient is a second-degree polynomial. Hence, the quotient is \(\color{blue}{3}\color{black}{}x^2+\color{blue}{7}\color{black}{}x−\color{blue}{4}\). The number in the bottom far-right corner, \(\color{red}{−9}\), is the remainder. Thus, the answer, written with the quotient and remainder, is \[\color{blue}{3}\color{black}{}x^2+\color{blue}{7}\color{black}{}x-\color{blue}{4}\color{black}{}-\dfrac{\color{red}{9}}{\color{black}{x-4}}\nonumber\]
If we compare this result to the result we obtained in Example 6.6.4 , we can see that they are identical. Whether we divided using polynomial division or synthetic division, we obtained the same result.
In general, synthetic division is the tool of choice for dividing polynomials by divisors of the form \(x−c\). It is important to note that it works only for these kinds of divisors. Good old-fashioned polynomial long division for divisors of degree larger than \(1\) should be used in other cases. Also, take note that when a polynomial (of degree at least \(1\)) is divided by \(x − c\), the result is a polynomial of exactly one less degree.
Polynomial Division Homework
Divide.
\(\dfrac{20x^4+x^3+2x^2}{4x^3}\)
\(\dfrac{20n^4+n^3+40n^2}{10n}\)
\(\dfrac{12x^4+24x^3+3x^2}{6x}\)
\(\dfrac{10n^4+50n^3+2n^2}{10n^2}\)
\(\dfrac{x^2-2x-71}{x+8}\)
\(\dfrac{n^2+13n+32}{n+5}\)
\(\dfrac{v^2-2v-89}{v-10}\)
\(\dfrac{a^2-4a-38}{a-8}\)
\(\dfrac{45p^2+56p+19}{9p+4}\)
\(\dfrac{10x^2-32x+9}{10x-2}\)
\(\dfrac{4r^2-r-1}{4r+3}\)
\(\dfrac{n^2-4}{n-2}\)
\(\dfrac{27b^2+87b+35}{3b+8}\)
\(\dfrac{4x^2-33x+28}{4x-5}\)
\(\dfrac{a^3+15a^2+49a-55}{a+7}\)
\(\dfrac{x^3-26x-41}{x+4}\)
\(\dfrac{3n^2+9n^2-64n-68}{n+6}\)
\(\dfrac{x^3-46x+22}{x+7}\)
\(\dfrac{9p^3+45p^2+27p-5}{9p+9}\)
\(\dfrac{r^3-r^2-16r+8}{r-4}\)
\(\dfrac{12n^3+12n^2-15n-4}{2n+3}\)
\(\dfrac{4v^3-21v^2+6v+19}{4v+3}\)
\(\dfrac{5x^4+45x^2+4x^2}{9x}\)
\(\dfrac{3k^3+4k^2+2k}{8k}\)
\(\dfrac{5p^4+16p^3+16p^2}{4p}\)
\(\dfrac{3m^4+18m^3+27m^2}{9m^2}\)
\(\dfrac{r^2-3r-53}{r-9}\)
\(\dfrac{b^2-10b+16}{b-7}\)
\(\dfrac{x^2+4x-26}{x+7}\)
\(\dfrac{x^2-10x+22}{x-4}\)
\(\dfrac{48k^2-70k+16}{6k-2}\)
\(\dfrac{n^2+7n+15}{n+4}\)
\(\dfrac{3m^2+9m-9}{3m-3}\)
\(\dfrac{2x^2-5x-8}{2x+3}\)
\(\dfrac{3v^2-32}{3v-9}\)
\(\dfrac{4n^2-23n-38}{4n+5}\)
\(\dfrac{8k^3-66k^2+12k+37}{k-8}\)
\(\dfrac{x^3-16x^2+71x-56}{x-8}\)
\(\dfrac{k^3-4k^2-6k+4}{k-1}\)
\(\dfrac{2n^3+21n^2+25n}{2n+3}\)
\(\dfrac{8m^3-57m^2+42}{8m+7}\)
\(\dfrac{2x^3+12x^2+4x-37}{2x+6}\)
\(\dfrac{24b^3-38b^2+29b-60}{4b-7}\)
Perform the indicated operations given the set of functions.
Let \(f(x) = x^3 − 2x^2 − 4x − 5\) and \(g(x) = x + 2\), find \((f\div g)(x)\).
Let \(f(n) = 3n + 5\) and \(k(n) = n^2 + 5\), find \((k\div f)(n)\).
Use synthetic division to divide.
\(x^3+4x^2+4x+6\) by \(x+1\)
\(x^4+4x^3-28x^2+26x-17\) by \(x-3\)
\(x^3+3x^2-2x+5\) by \(x-1\)
\(x^4+5x^3+11x^2+13x-2\) by \(x+2\)
\(x^3+x^2-4x-1\) by \(x-2\)
\(x^4-4x^3-6x^2-4x-8\) by \(x+1\)
\(x^3+4x^2-6x-5\) by \(x-2\)
\(x^4+10x^3+11x^2-15x-14\) by \(x+2\)
\(x^3+11x^2+26x+12\) by \(x+3\)
\(x^4-4x^3-3x^2+6x+19\) by \(x-2\)