maths.free › Precalculus › 3. Polynomial and Rational Functions › Dividing Polynomials

Dividing Polynomials

Use long division to divide polynomials.

Using Long Division to Divide Polynomials

We are familiar with the long division algorithm for ordinary arithmetic. We begin by dividing into the digits of the dividend that have the greatest place value. We divide, multiply, subtract, include the digit in the next place value position, and repeat. For example, let’s divide 178 by 3 using long division.

Another way to look at the solution is as a sum of parts. This should look familiar, since it is the same method used to check division in elementary arithmetic.

\[\begin{array}{l}\text{dividend = }(\text{divisor }⋅\text{ quotient})\text{ + remainder} \\ 178=(3⋅59)+1 \\ =177+1 \\ =178\end{array}\]

We call this the Division Algorithm and will discuss it more formally after looking at an example.

Division of polynomials that contain more than one term has similarities to long division of whole numbers. We can write a polynomial dividend as the product of the divisor and the quotient added to the remainder. The terms of the polynomial division correspond to the digits (and place values) of the whole number division. This method allows us to divide two polynomials. For example, if we were to divide \(2{x}^{3}-3{x}^{2}+4x+5\) by \(x+2\) using the long division algorithm, it would look like this:

We have found

\[\frac{2{x}^{3}-3{x}^{2}+4x+5}{x+2}=2{x}^{2}-7x+18-\frac{31}{x+2}\]

\[2{x}^{3}-3{x}^{2}+4x+5=(x+2)(2{x}^{2}-7x+18)-31\]

We can identify the dividend, the divisor, the quotient, and the remainder.

Condensed — the full section is in OpenStax Precalculus 2e.

Using Synthetic Division to Divide Polynomials

As we’ve seen, long division of polynomials can involve many steps and be quite cumbersome. Synthetic division is a shorthand method of dividing polynomials for the special case of dividing by a linear factor whose leading coefficient is 1.

To illustrate the process, recall the example at the beginning of the section.

Divide \(2{x}^{3}-3{x}^{2}+4x+5\) by \(x+2\) using the long division algorithm.

The final form of the process looked like this:

There is a lot of repetition in the table. If we don’t write the variables but, instead, line up their coefficients in columns under the division sign and also eliminate the partial products, we already have a simpler version of the entire problem.

Synthetic division carries this simplification even a few more steps. Collapse the table by moving each of the rows up to fill any vacant spots. Also, instead of dividing by 2, as we would in division of whole numbers, then multiplying and subtracting the middle product, we change the sign of the “divisor” to –2, multiply and add. The process starts by bringing down the leading coefficient.

Example

Try it.

Use synthetic division to divide \(5{x}^{2}-3x-36\) by \(x-3.\)

Solution

Begin by setting up the synthetic division. Write \(k\) and the coefficients.

Bring down the lead coefficient. Multiply the lead coefficient by \(k.\)

Continue by adding the numbers in the second column. Multiply the resulting number by \(k.\) Write the result in the next column. Then add the numbers in the third column.

The result is \(5x+12.\) The remainder is 0. So \(x-3\) is a factor of the original polynomial.

Example

Try it.

Use synthetic division to divide \(-9{x}^{4}+10{x}^{3}+7{x}^{2}-6\) by \(x-1.\)

Solution

Notice there is no x-term. We will use a zero as the coefficient for that term.

The result is \(-9{x}^{3}+{x}^{2}+8x+8+\frac{2}{x-1}.\)

Condensed — the full section is in OpenStax Precalculus 2e.

Using Polynomial Division to Solve Application Problems

Polynomial division can be used to solve a variety of application problems involving expressions for area and volume. We looked at an application at the beginning of this section. Now we will solve that problem in the following example.

Example

Try it.

The volume of a rectangular solid is given by the polynomial \(3{x}^{4}-3{x}^{3}-33{x}^{2}+54x.\) The length of the solid is given by \(3x\) and the width is given by \(x-2.\) Find the height of the solid.

Solution

There are a few ways to approach this problem. We need to divide the expression for the volume of the solid by the expressions for the length and width. Let us create a sketch as in .

We can now write an equation by substituting the known values into the formula for the volume of a rectangular solid.

\[\begin{array}{l}\ V=l⋅w⋅h \\ 3{x}^{4}-3{x}^{3}-33{x}^{2}+54x=3x⋅(x-2)⋅h\end{array}\]

To solve for \(h,\) first divide both sides by \(3x.\)

\[\begin{array}{l}\frac{3x⋅(x-2)⋅h}{3x}=\frac{3{x}^{4}-3{x}^{3}-33{x}^{2}+54x}{3x} \\ (x-2)h={x}^{3}-{x}^{2}-11x+18\end{array}\]

Now solve for \(h\) using synthetic division.

\[h=\frac{{x}^{3}-{x}^{2}-11x+18}{x-2}\]\[\begin{array}{lllllll} \\ 2\begin{array}{llll}1 & -1 & -11 & 18 \\ & 2 & 2 & -18\end{array} \\ \begin{array}{llll}\ 1 & \ 1 & \ -9 & \ 0\end{array}\end{array}\]

The quotient is \({x}^{2}+x-9\) and the remainder is 0. The height of the solid is \({x}^{2}+x-9.\)

Key Concepts

Polynomial long division can be used to divide a polynomial by any polynomial with equal or lower degree. See and .
The Division Algorithm tells us that a polynomial dividend can be written as the product of the divisor and the quotient added to the remainder.
Synthetic division is a shortcut that can be used to divide a polynomial by a binomial in the form \(x-k.\) See , , and .
Polynomial division can be used to solve application problems, including area and volume. See .

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

Divide \(5{x}^{2}+3x-2\) by \(x+1.\)

i

The quotient is \(5x-2.\) The remainder is 0. We write the result as
\[\frac{5{x}^{2}+3x-2}{x+1}=5x-2\]
or
\[5{x}^{2}+3x-2=(x+1)(5x-2)\]
Divide \(6{x}^{3}+11{x}^{2}-31x+15\) by \(3x-2.\)

i

There is a remainder of 1. We can express the result as:
\[\frac{6{x}^{3}+11{x}^{2}-31x+15}{3x-2}=2{x}^{2}+5x-7+\frac{1}{3x-2}\]
Divide \(16{x}^{3}-12{x}^{2}+20x-3\) by \(4x+5.\)

i

\(4{x}^{2}-8x+15-\frac{78}{4x+5}\)
Use synthetic division to divide \(5{x}^{2}-3x-36\) by \(x-3.\)

i

Begin by setting up the synthetic division. Write \(k\) and the coefficients.
Bring down the lead coefficient. Multiply the lead coefficient by \(k.\)
Continue by adding the numbers in the second column. Multiply the resulting number by \(k.\) Write the result in the next column. Then add the numbers in the third column.
The result is \(5x+12.\) The remainder is 0. So \(x-3\) is a factor of the original polynomial.
Use synthetic division to divide \(4{x}^{3}+10{x}^{2}-6x-20\) by \(x+2.\)

i

The binomial divisor is \(x+2\) so \(k=-2.\) Add each column, multiply the result by –2, and repeat until the last column is reached.
The result is \(4{x}^{2}+2x-10.\) The remainder is 0. Thus, \(x+2\) is a factor of \(4{x}^{3}+10{x}^{2}-6x-20.\)
Use synthetic division to divide \(-9{x}^{4}+10{x}^{3}+7{x}^{2}-6\) by \(x-1.\)

i

Notice there is no x-term. We will use a zero as the coefficient for that term.
The result is \(-9{x}^{3}+{x}^{2}+8x+8+\frac{2}{x-1}.\)
Use synthetic division to divide \(3{x}^{4}+18{x}^{3}-3x+40\) by \(x+7.\)

i

\(3{x}^{3}-3{x}^{2}+21x-150+\frac{1,090}{x+7}\)
The volume of a rectangular solid is given by the polynomial \(3{x}^{4}-3{x}^{3}-33{x}^{2}+54x.\) The length of the solid is given by \(3x\) and the width is given by \(x-2.\) Find the height of the solid.

i

There are a few ways to approach this problem. We need to divide the expression for the volume of the solid by the expressions for the length and width. Let us create a sketch as in .
We can now write an equation by substituting the known values into the formula for the volume of a rectangular solid.
\[\begin{array}{l}\ V=l⋅w⋅h \\ 3{x}^{4}-3{x}^{3}-33{x}^{2}+54x=3x⋅(x-2)⋅h\end{array}\]
To solve for \(h,\) first divide both sides by \(3x.\)
\[\begin{array}{l}\frac{3x⋅(x-2)⋅h}{3x}=\frac{3{x}^{4}-3{x}^{3}-33{x}^{2}+54x}{3x} \\ (x-2)h={x}^{3}-{x}^{2}-11x+18\end{array}\]
Now solve for \(h\) using synthetic division.
\[h=\frac{{x}^{3}-{x}^{2}-11x+18}{x-2}\]\[\begin{array}{lllllll} \\ 2\begin{array}{llll}1 & -1 & -11 & 18 \\ & 2 & 2 & -18\end{array} \\ \begin{array}{llll}\ 1 & \ 1 & \ -9 & \ 0\end{array}\end{array}\]
The quotient is \({x}^{2}+x-9\) and the remainder is 0. The height of the solid is \({x}^{2}+x-9.\)
The area of a rectangle is given by \(3{x}^{3}+14{x}^{2}-23x+6.\) The width of the rectangle is given by \(x+6.\) Find an expression for the length of the rectangle.

i

\(3{x}^{2}-4x+1\)
If division of a polynomial by a binomial results in a remainder of zero, what can be conclude?

i

The binomial is a factor of the polynomial.
If a polynomial of degree \(n\) is divided by a binomial of degree 1, what is the degree of the quotient?
\(({x}^{2}+5x-1)\div (x-1)\)

i

\(x+6+\frac{5}{x-1}\), quotient: \(x+6\), remainder: \(5\)
\((2{x}^{2}-9x-5)\div (x-5)\)
\((3{x}^{2}+23x+14)\div (x+7)\)

i

\(3x+2\), quotient: \(3x+2\), remainder: \(0\)
\((4{x}^{2}-10x+6)\div (4x+2)\)
\((6{x}^{2}-25x-25)\div (6x+5)\)

i

\(x-5\), quotient: \(x-5\), remainder: \(0\)
\((-{x}^{2}-1)\div (x+1)\)
\((2{x}^{2}-3x+2)\div (x+2)\)

i

\(2x-7+\frac{16}{x+2}\), quotient: \(2x-7\), remainder: \(16\)
\(({x}^{3}-126)\div (x-5)\)
\((3{x}^{2}-5x+4)\div (3x+1)\)

i

\(x-2+\frac{6}{3x+1}\), quotient: \(x-2\), remainder: \(6\)
\(({x}^{3}-3{x}^{2}+5x-6)\div (x-2)\)
\((2{x}^{3}+3{x}^{2}-4x+15)\div (x+3)\)

i

\(2{x}^{2}-3x+5\), quotient: \(2{x}^{2}-3x+5\), remainder: \(0\)
\((3{x}^{3}-2{x}^{2}+x-4)\div (x+3)\)
\((2{x}^{3}-6{x}^{2}-7x+6)\div (x-4)\)

i

\(2{x}^{2}+2x+1+\frac{10}{x-4}\)
\((6{x}^{3}-10{x}^{2}-7x-15)\div (x+1)\)
\((4{x}^{3}-12{x}^{2}-5x-1)\div (2x+1)\)

i

\(2{x}^{2}-7x+1-\frac{2}{2x+1}\)
\((9{x}^{3}-9{x}^{2}+18x+5)\div (3x-1)\)
\((3{x}^{3}-2{x}^{2}+x-4)\div (x+3)\)

i

\(3{x}^{2}-11x+34-\frac{106}{x+3}\)
\((-6{x}^{3}+{x}^{2}-4)\div (2x-3)\)
\((2{x}^{3}+7{x}^{2}-13x-3)\div (2x-3)\)

i

\({x}^{2}+5x+1\)
\((3{x}^{3}-5{x}^{2}+2x+3)\div (x+2)\)
\((4{x}^{3}-5{x}^{2}+13)\div (x+4)\)

i

\(4{x}^{2}-21x+84-\frac{323}{x+4}\)
\(({x}^{3}-3x+2)\div (x+2)\)
\(({x}^{3}-21{x}^{2}+147x-343)\div (x-7)\)

i

\({x}^{2}-14x+49\)
\(({x}^{3}-15{x}^{2}+75x-125)\div (x-5)\)
\((9{x}^{3}-x+2)\div (3x-1)\)

i

\(3{x}^{2}+x+\frac{2}{3x-1}\)
\((6{x}^{3}-{x}^{2}+5x+2)\div (3x+1)\)
\(({x}^{4}+{x}^{3}-3{x}^{2}-2x+1)\div (x+1)\)

i

\({x}^{3}-3x+1\)
\(({x}^{4}-3{x}^{2}+1)\div (x-1)\)
\(({x}^{4}+2{x}^{3}-3{x}^{2}+2x+6)\div (x+3)\)

i

\({x}^{3}-{x}^{2}+2\)

Symbols used here

|x|

absolute value / modulus
Distance from zero: |−3| = 3. For a complex number, distance from the origin.

\binom{n}{k}

binomial coefficient, "n choose k"
Number of k-element subsets of n things: n!/(k!(n−k)!).

\pi

pi
Ratio of a circle's circumference to its diameter, 3.14159…

Euler's number
2.71828…, the base whose exponential is its own derivative.

imaginary unit
i² = −1.

\theta

theta
The usual name for an angle.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\mathbf{v},\ \vec{v}

vector
A quantity with magnitude and direction; a column of numbers.

How to: Dividing Polynomials

Use long division to divide polynomials.
Use synthetic division to divide polynomials.
Set up the division problem.
Determine the first term of the quotient by dividing the leading term of the dividend by the leading term of the divisor.
Multiply the answer by the divisor and write it below the like terms of the dividend.
Subtract the bottom
Bring down the next term of the dividend.
Repeat steps 2–5 until reaching the last term of the dividend.

Questions people ask

What is a function, really?

A rule that assigns exactly one output to each input. The vertical-line test on a graph is the same idea: no input may have two outputs.

Why do we need complex numbers?

Because x² + 1 = 0 has no real solution, and allowing one new number i with i² = −1 makes every polynomial equation solvable. They then turn out to describe rotation, waves and alternating current more naturally than real numbers do.

Kuri Gukoresha

Parts of this page are adapted from OpenStax Precalculus 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

in Precalculus

Complex numbers Polynomial functions Rational functions Sequences and series The binomial theorem Conic sections Vectors Exponential and logarithmic functions Polynomial division and the remainder theorem Parametric equations and polar coordinates