In the next two sections, we will be learning two ways to divide polynomials. These techniques can help you find the zeros of a polynomial that is not factorable over the integers.

recall long division

A handy algebraic technique involves rewriting a quotient of polynomials as a polynomial of reduced degree plus a remainder quotient

[latex]\dfrac{p(x)}{q(x)}=h(x)+\dfrac{r(x)}{q(x)}[/latex].

Or equivalently by multiplying both sides by [latex]q(x)[/latex],

[latex]p(x)=q(x)h(x) + r(x)[/latex]

This method is akin to rewriting an “improper” fraction such as

[latex]\dfrac{17}{3} = 5 + \dfrac{2}{3}[/latex].

Or equivalently,

[latex]17=3\cdot 5 + 2[/latex].

The process used to accomplish this is similar to long division of whole numbers, so if you haven’t performed long division for awhile, take a moment to refresh before reading this section.

more

Recall how you can use long division to divide two whole numbers, say [latex]900[/latex] divided by [latex]37[/latex].

First, you would think about how many [latex]37s[/latex] are in [latex]90[/latex], as [latex]9[/latex] is too small. (Note: you could also think, how many [latex]40s[/latex] are there in [latex]90[/latex].)

There are two [latex]37s[/latex] in [latex]90[/latex], so write [latex]2[/latex] above the last digit of [latex]90[/latex]. Two [latex]37s[/latex] is [latex]74[/latex]; write that product below the [latex]90[/latex].

Subtract: [latex]90–74[/latex] is [latex]16[/latex]. (If the result is larger than the divisor, [latex]37[/latex], then you need to use a larger number for the quotient.)

Bring down the next digit [latex](0)[/latex] and consider how many [latex]37s[/latex] are in [latex]160[/latex].

There are four [latex]37s[/latex] in [latex]160[/latex], so write the [latex]4[/latex] next to the two in the quotient. Four [latex]37s[/latex] is [latex]148[/latex]; write that product below the [latex]160[/latex].

Subtract: [latex]160–148[/latex] is [latex]12[/latex]. This is less than [latex]37[/latex] so the [latex]4[/latex] is correct. Since there are no more digits in the dividend to bring down, you are done.

The final answer is [latex]24[/latex] R[latex]12[/latex], or [latex]24\frac{12}{37}[/latex]. You can check this by multiplying the quotient (without the remainder) by the divisor, and then adding in the remainder. The result should be the dividend:

[latex]24\cdot37+12=888+12=900[/latex]

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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.

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

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, divisor, quotient, and remainder.

Writing the result in this manner illustrates the Division Algorithm.

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].

Show Solution

There is a remainder of 1. We can express the result as:

[latex]\frac{6{x}^{3}+11{x}^{2}-31x+15}{3x - 2}=2{x}^{2}+5x - 7+\frac{1}{3x - 2}[/latex]

Analysis of the Solution

We can check our work by using the Division Algorithm to rewrite the solution then multiplying.

[latex]\left(3x - 2\right)\left(2{x}^{2}+5x - 7\right)+1=6{x}^{3}+11{x}^{2}-31x+15[/latex]

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