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Study Guides > College Algebra: Co-requisite Course

Division of Polynomials

Learning Objectives

  • Divide a polynomial by a monomial
  • Long division of polynomials
  • Applications of polynomial division
The exterior of the Lincoln Memorial in Washington, D.C., is a large rectangular solid with length 61.5 meters (m), width 40 m, and height 30 m.[footnote]National Park Service. "Lincoln Memorial Building Statistics." http://www.nps.gov/linc/historyculture/lincoln-memorial-building-statistics.htm. Accessed 4/3/2014[/footnote] We can easily find the volume using elementary geometry.

[latex]\begin{array}{l}V=l\cdot w\cdot h\hfill \\ \text{ }=61.5\cdot 40\cdot 30\hfill \\ \text{ }=73,800\hfill \end{array}[/latex]

So the volume is 73,800 cubic meters [latex]\left(\text{m}{^3} \right)[/latex]. Suppose we knew the volume, length, and width. We could divide to find the height.

[latex]\begin{array}{l}h=\frac{V}{l\cdot w}\hfill \\ \text{ }=\frac{73,800}{61.5\cdot 40}\hfill \\ \text{ }=30\hfill \end{array}[/latex]

As we can confirm from the dimensions above, the height is 30 m. We can use similar methods to find any of the missing dimensions. We can also use the same method if any or all of the measurements contain variable expressions. For example, suppose the volume of a rectangular solid is given by the polynomial [latex]3{x}^{4}-3{x}^{3}-33{x}^{2}+54x[/latex]. The length of the solid is given by 3x; the width is given by [latex]x - 2[/latex]. To find the height of the solid, we can use polynomial division, which is the focus of this section.
Lincoln Memorial. Lincoln Memorial, Washington, D.C. (credit: Ron Cogswell, Flickr)

Divide a polynomial by a monomial

Division of polynomials isn’t much different from division of numbers. In the exponential section, you were asked to simplify expressions such as: [latex]\displaystyle\frac{{{a}^{2}}{{({{a}^{5}})}^{3}}}{8{{a}^{8}}}[/latex]. This expression is the division of two monomials. To simplify it, we divided the coefficients and then divided the variables. In this section we will add another layer to this idea by dividing polynomials by monomials, and by binomials. The distributive property states that you can distribute a factor that is being multiplied by a sum or difference, and likewise you can distribute a divisor that is being divided into a sum or difference. In this example, you can add all the terms in the numerator, then divide by 2.

[latex]\displaystyle \frac{\text{dividend}\rightarrow}{\text{divisor}\rightarrow}\,\,\,\,\,\, \frac{8+4+10}{2}=\frac{22}{2}=11[/latex]

Or you can first divide each term by 2, then simplify the result.

[latex]\displaystyle \frac{8}{2}+\frac{4}{2}+\frac{10}{2}=4+2+5=11[/latex]

Either way gives you the same result. The second way is helpful when you can't combine like terms in the numerator.  Let’s try something similar with a binomial.

Example

Divide. [latex]\displaystyle \frac{9a^3+6a}{3a^2}[/latex]

Answer: Distribute [latex]3a^2[/latex] over the polynomial by dividing each term by [latex]3a^2[/latex] [latex-display]\displaystyle \frac{9a^3}{3a^2}+\frac{6a}{3a^2}[/latex-display] Divide each term, a monomial divided by another monomial. [latex-display]\displaystyle \begin{array}{c}3a^{3-2}+2a^{1-2}\\\text{ }\\=3a^{1}+2a^{-1}\\\text{ }\\=3a+2a^{-1}\end{array}[/latex-display] Rewrite [latex]a^{-1}[/latex] with positive exponents, as a matter of convention. [latex-display]\displaystyle 3a+2a^{-1}=3a+\frac{2}{a}[/latex-display]

Answer

[latex-display]\displaystyle \frac{9a^3+6a}{3a^2}=3a+\frac{2}{a}[/latex-display]

In the next example, you will see that the same ideas apply when you are dividing a trinomial by a monomial. You can distribute the divisor to each term in the trinomial and simplify using the rules for exponents. As we have throughout the course, simplifying with exponents includes rewriting negative exponents as positive. Pay attention to the signs of the terms in the next example, we will divide by a negative monomial.

Example

Divide. [latex]\displaystyle \frac{27{{y}^{4}}+6{{y}^{2}}-18}{-6y}[/latex]

Answer: Divide each term in the polynomial by the monomial.

[latex]\displaystyle \frac{27{{y}^{4}}}{-6y}+\frac{6{{y}^{2}}}{-6y}-\frac{18}{-6y}[/latex]

Note how the term[latex]\displaystyle -\frac{18}{-6y}[/latex] does not have a y in the numerator, so division is only applied to the numbers [latex]18, -6[/latex]. Also, 27 doesn't divide nicely by [latex]-6[/latex], so we are left with a fraction as the coefficient on the [latex]y^3[/latex] term. Simplify.

[latex]\displaystyle -\frac{9}{2}{{y}^{3}}-y+\frac{3}{y}[/latex]

Answer

[latex-display]\displaystyle \frac{27{{y}^{4}}+6{{y}^{2}}-18}{-6y}=-\frac{9}{2}{{y}^{3}}-y+\frac{3}{y}[/latex-display]

The same principle will apply for polynomials with more than one variable.

ExAMPLE

Divide. [latex]\displaystyle \frac{18x^3y-36xy^2+12x}{-3xy}[/latex]

Answer: Divide each term by the divisor. Be careful with the signs!

[latex]\displaystyle \frac{18x^3y}{-3xy}-\frac{36xy^2}{-3xy}+\frac{12x}{-3xy} [/latex]

Simplify, using the rules of exponents.

[latex]\displaystyle -6x^2+12y-\frac{4}{y}[/latex]

Answer

[latex-display]\displaystyle -6x^2+12y-\frac{4}{y}[/latex-display]

Notice, that in the examples above, the answer was a rational expression.

Try It

[ohm_question]38831[/ohm_question]
   

Polynomial Long Division

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. Long Division. Step 1, 5 times 3 equals 15 and 17 minus 15 equals 2. Step 2: Bring down the 8. Step 3: 9 times 3 equals 27 and 28 minus 27 equals 1. Answer: 59 with a remainder of 1 or 59 and one-third. 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.

[latex]\begin{array}{l}\text{dividend = }\left(\text{divisor }\cdot \text{ quotient}\right)\text{ + remainder}\hfill \\ 178=\left(3\cdot 59\right)+1\hfill \\ =177+1\hfill \\ =178\hfill \end{array}[/latex]

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 [latex]2{x}^{3}-3{x}^{2}+4x+5[/latex] by [latex]x+2[/latex] using the long division algorithm, it would look like this: Set up the division problem. 2x cubed divided by x is 2x squared. Multiply the sum of x and 2 by 2x squared. Subtract. Then bring down the next term. Negative 7x squared divided by x is negative 7x. Multiply the sum of x and 2 by negative 7x. Subtract, then bring down the next term. 18x divided by x is 18. Multiply the sum of x and 2 by 18. Subtract. We have found

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

or

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

We can identify the dividend, the divisor, the quotient, and the remainder. The dividend is 2x cubed minus 3x squared plus 4x plus 5. The divisor is x plus 2. The quotient is 2x squared minus 7x plus 18. The remainder is negative 31. Writing the result in this manner illustrates the Division Algorithm.

A General Note: The Division Algorithm

The Division Algorithm states that, given a polynomial dividend [latex]f\left(x\right)[/latex] and a non-zero polynomial divisor [latex]d\left(x\right)[/latex] where the degree of [latex]d\left(x\right)[/latex] is less than or equal to the degree of [latex]f\left(x\right)[/latex], there exist unique polynomials [latex]q\left(x\right)[/latex] and [latex]r\left(x\right)[/latex] such that

[latex]f\left(x\right)=d\left(x\right)q\left(x\right)+r\left(x\right)[/latex]

[latex]q\left(x\right)[/latex] is the quotient and [latex]r\left(x\right)[/latex] is the remainder. The remainder is either equal to zero or has degree strictly less than [latex]d\left(x\right)[/latex]. If [latex]r\left(x\right)=0[/latex], then [latex]d\left(x\right)[/latex] divides evenly into [latex]f\left(x\right)[/latex]. This means that, in this case, both [latex]d\left(x\right)[/latex] and [latex]q\left(x\right)[/latex] are factors of [latex]f\left(x\right)[/latex].

How To: Given a polynomial and a binomial, use long division to divide the polynomial by the binomial.

  1. Set up the division problem.
  2. Determine the first term of the quotient by dividing the leading term of the dividend by the leading term of the divisor.
  3. Multiply the answer by the divisor and write it below the like terms of the dividend.
  4. Subtract the bottom binomial from the top binomial.
  5. Bring down the next term of the dividend.
  6. Repeat steps 2–5 until reaching the last term of the dividend.
  7. If the remainder is non-zero, express as a fraction using the divisor as the denominator.

Example: Using Long Division to Divide a Second-Degree Polynomial

Divide [latex]5{x}^{2}+3x - 2[/latex] by [latex]x+1[/latex].

Answer: Set up the division problem. 5x squared divided by x is 5x. Multiply x plus 1 by 5x. Subtract. Bring down the next term. Negative 2x divded by x is negative 2. Multiply x + 1 by negative 2. Subtract.The quotient is [latex]5x - 2[/latex]. The remainder is 0. We write the result as

[latex]\frac{5{x}^{2}+3x - 2}{x+1}=5x - 2[/latex]

or

[latex]5{x}^{2}+3x - 2=\left(x+1\right)\left(5x - 2\right)[/latex]

Analysis of the Solution

This division problem had a remainder of 0. This tells us that the dividend is divided evenly by the divisor, and that the divisor is a factor of the dividend.

Example: Using Long Division to Divide a Third-Degree Polynomial

Divide [latex]6{x}^{3}+11{x}^{2}-31x+15[/latex] by [latex]3x - 2[/latex].

Answer: 6x cubed divided by 3x is 2x squared. Multiply the sum of x and 2 by 2x squared. Subtract. Bring down the next term. 15x squared divided by 3x is 5x. Multiply 3x minus 2 by 5x. Subtract. Bring down the next term. Negative 21x divided by 3x is negative 7. Multiply 3x minus 2 by negative 7. Subtract. The remainder is 1. There is a remainder of 1. We can express the result as: [latex-display]\frac{6{x}^{3}+11{x}^{2}-31x+15}{3x - 2}=2{x}^{2}+5x - 7+\frac{1}{3x - 2}[/latex-display]

Analysis of the Solution

We can check our work by using the Division Algorithm to rewrite the solution. Then multiply. [latex-display]\left(3x - 2\right)\left(2{x}^{2}+5x - 7\right)+1=6{x}^{3}+11{x}^{2}-31x+15[/latex-display] Notice, as we write our result,
  • the dividend is [latex]6{x}^{3}+11{x}^{2}-31x+15[/latex]
  • the divisor is [latex]3x - 2[/latex]
  • the quotient is [latex]2{x}^{2}+5x - 7[/latex]
  • the remainder is 1

Try It

Divide [latex]16{x}^{3}-12{x}^{2}+20x - 3[/latex] by [latex]4x+5[/latex].

Answer: [latex]4{x}^{2}-8x+15-\frac{78}{4x+5}[/latex]

 

Example

Divide: [latex]\frac{\left(x^{3}–6x–10\right)}{\left(x–3\right)}[/latex]

Answer: In setting up this problem, notice that there is an [latex]x^{3}[/latex] term but no [latex]x^{2}[/latex] term. Add [latex]0x^{2}[/latex] as a “place holder” for this term. (Since 0 times anything is 0, you’re not changing the value of the dividend.) Focus on the first terms again: how many x’s are there in [latex]x^{3}[/latex]? Since [latex] \frac{{{x}^{3}}}{x}=x^{2}[/latex], put [latex]x^{2}[/latex] in the quotient. Multiply [latex]x^{2}\left(x–3\right)=x^{3}–3x^{2}[/latex], write this underneath the dividend, and prepare to subtract. Rewrite the subtraction using the opposite of the expression [latex]x^{3}-3x^{2}[/latex]. Then add. Bring down the rest of the expression in the dividend. It’s helpful to bring down all of the remaining terms. Now, repeat the process with the remaining expression, [latex]3x^{2}-6x–10[/latex], as the dividend. Remember to watch the signs! How many x’s are there in 3x? Since there are 3, multiply [latex]3\left(x–3\right)=3x–9[/latex], write this underneath the dividend, and prepare to subtract. Continue until the degree of the remainder is less than the degree of the divisor. In this case the degree of the remainder, [latex]-1[/latex], is 0, which is less than the degree of [latex]x-3[/latex], which is 1. Also notice that you have brought down all the terms in the dividend, and that the quotient extends to the right edge of the dividend. These are other ways to check whether you have completed the problem. You can write the remainder using the symbol R, or as a fraction added to the rest of the quotient with the remainder in the numerator and the divisor in the denominator. In this case, since the remainder is negative, you can also subtract the opposite.

Answer

[latex-display]\begin{array}{r}{\left(x^{3}–6x–10\right)}{\left(x–3\right)}=x^{2}+3x+3+R-1,\\x^{2}+3x+3+\frac{-1}{x-3}, \text{ or }\\x^{2}+3x+3-\frac{1}{x-3}\end{array}[/latex-display]

Check the result:

[latex]\left(x–3\right)\left(x^{2}+3x+3\right)\,\,\,=\,\,\,x\left(x^{2}+3x+3\right)–3\left(x^{2}+3x+3\right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\,\,\,x^{3}+3x^{2}+3x–3x^{2}–9x–9\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\,\,\,x^{3}–6x–9\\\,\,\,\,\,\,\,\,x^{3}–6x–9+\left(-1\right)\,\,\,=\,\,\,x^{3}–6x–10[/latex]

In the video that follows, we show another example of dividing a degree three trinomial by a binomial, not the "missing" term and how we work with it. https://youtu.be/Rxds7Q_UTeo  

Application of Polynomial Division

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: Using Polynomial Division in an Application Problem

The volume of a rectangular solid is given by the polynomial [latex]3{x}^{4}-3{x}^{3}-33{x}^{2}+54x[/latex]. The length of the solid is given by 3x and the width is given by x – 2. Find the height of the solid.

Answer: 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. Graph of f(x)=4x^3+10x^2-6x-20 with a close up on x+2. We can now write an equation by substituting the known values into the formula for the volume of a rectangular solid.

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

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

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

Now solve for h using synthetic division.

[latex]h=\frac{{x}^{3}-{x}^{2}-11x+18}{x - 2}[/latex]

Synthetic division with 2 as the divisor and {1, -1, -11, 18} as the quotient. The result is {1, 1, -9, 0} The quotient is [latex]{x}^{2}+x - 9[/latex] and the remainder is 0. The height of the solid is [latex]{x}^{2}+x - 9[/latex].

Try It

The area of a rectangle is given by [latex]3{x}^{3}+14{x}^{2}-23x+6[/latex]. The width of the rectangle is given by x + 6. Find an expression for the length of the rectangle.

Answer: [latex]3{x}^{2}-4x+1[/latex]

 

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  • College Algebra. Provided by: OpenStax Authored by: Abramson, Jay et al.. Located at: https://openstax.org/books/college-algebra/pages/1-introduction-to-prerequisites. License: CC BY: Attribution. License terms: Download for free at http://cnx.org/contents/[email protected].
  • Question ID 29482, 29483. Authored by: McClure, Caren. License: CC BY: Attribution. License terms: IMathAS Community License CC-BY + GPL.
  • Divide a Degree 3 Polynomial by a Degree 1 Polynomial (Long Division with Missing Term). Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.
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