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Mathematics LibreTexts

4.4: Polynomial Division

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Learning Objectives
  • Divide polynomials using long division
  • Divide polynomials using synthetic division
  • Apply the factor and remainder theorems to polynomials

Are You Ready?

Try these questions prior to beginning this section to help determine if you are set up for success:

  1. Simplify the expression:
    1. (4x2+2x+1)(x4)+1
    2. 15z418z3+24z29z3z
Answer
    1. 4x314x27x3
    2. 5z36z2+8z3

If you missed this problem or feel you could use more practice, review [ 2.6: Multiplying Polynomial Expressions and 2.13: Simplifying, Multiplying, and Dividing Rational Expressions]

Suppose we wish to find the x-intercepts of f(x)=x3+4x25x14. Setting f(x)=0 results in the polynomial equation x3+4x25x14=0. Despite all of the factoring techniques we learned in Intermediate Algebra courses, this equation foils us at every turn since these factoring techniques do no work. If we graph f using a graphing calculator window [-5,5]x[-16,16], we get

3.2a.png

The graph suggests that the function has three x-intercepts, one of which is at x=2. It's easy to show that f(2)=0, but the other two x-intercepts seem to be less friendly. Even though we could use the 'Zero' command to find decimal approximations for these x-intercepts, we seek a method to find the exact values of the other x-intercepts. Based on our experience in the previous section, if there is an x-intercept at x=2, it seems that there should be a factor of (x2) for f(x). In other words, we should expect that x3+4x25x14=(x2)q(x), where q(x) is some other polynomial. How could we find such a q(x), if it even exists? The answer comes from our old friend, division.

We are familiar with long division from 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

clipboard_e6aedd04ae3536b4b0c37e94575316581.png

 

Step 1: 5×3=15 and 1715=2
Step 2: Bring down the 8
Step 3: 9×3=27 and 2827=1

Answer: 59R1 or 5913

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.

dividend=(divisorquotient)+remainder178=(359)+1=177+1=178

We can also state that 1783=59+13 which is the algebraic form of the mixed number 5913

We call this the Division Algorithm.

Long Division of Polynomials

Now, let's revisit the the polynomial f(x)=x3+4x25x14. In the last section we saw that we could write a polynomial as a product of factors, each corresponding to a x-intercept. If we know that there is an x-intercept at x=2 for f(x), then we might guess that the polynomial could be factored as x3+4x25x14=(x2) (something). To find that "something," we can use polynomial division.

Example 4.4.1

Divide x3+4x25x14 by x2.

Solution

Start by writing the problem out in long division form

x2\longdivx3+4x25x14

Now we divide the leading terms: x3÷x=x2. It is best to align it above the same-powered term in the dividend. Now, multiply that x2 by x2 and write the result below the dividend.

屏幕快照 2019-06-23 上午5.42.53.png

Again, divide the leading term of the remainder by the leading term of the divisor. 6x2÷x=6x. We add this to the result, multiply 6x by x2, and subtract.

屏幕快照 2019-06-23 上午5.43.34.png

This tells us x3+4x25x14 divided by x2 is x2+6x+7, with a remainder of zero. This also means that we can factor x3+4x25x14 as (x2)(x2+6x+7).

Now, if we want to find the other x-intercepts, we now solve x2+6x+7=0. Since this doesn't factor nicely we can complete the square or use the Quadratic Formula to get x=3±2. The point of this section is to focus on how to perform division. Before looking at more examples, let's note what we can expect when we divide polynomials.

The Division Algorithm for Polynomials

Suppose d(x) and p(x) are nonzero polynomials where the degree of p is greater than or equal to the degree of d. There exist two unique polynomials, q(x) and r(x), such that p(x)=d(x)q(x)+r(x),

or p(x)d(x)=q(x)+r(x)d(x)

where either r(x)=0 or the degree of r is strictly less than the degree of d.

The polynomial p is called the dividend; d is the divisor; q is the quotient; r is the remainder. If r(x)=0 then d is called a factor of p.

Because of the division, the remainder will either be zero, or a polynomial of lower degree than d(x). Because of this, if we divide a polynomial by a term of the form xc, then the remainder will be zero or a constant.

If p(x)=(xc)q(x)+r, then p(c)=(cc)q(c)+r=0+r=r, which establishes the Remainder Theorem.

The Remainder Theorem

If p(x) is a polynomial of degree 1 or greater and c is a real number, then when p(x) is divided by xc, the remainder is p(c).

If xc is a factor of the polynomial p, then p(x)=(xc)q(x) for some polynomial q. Then p(c)=(cc)q(c)=0, showing c is a zero of the polynomial. This shouldn’t surprise us - we already knew that if the polynomial factors it reveals the roots.

If p(c)=0, then the remainder theorem tells us that if p is divided by xc, then the remainder will be zero, which means xc is a factor of p.

The solutions to an equation p(x)=0 are called zeros of p(x). Zeros can be real numbers or complex numbers. As we have seen with quadratics, real zeros corresponding to x-coordinates of x-intercepts and complex zeros do not.

The Factor Theorem

If p(x) is a nonzero polynomial, then the real number c is a zero of p(x) if and only if xc is a factor of p(x).

Example 4.4.2

Divide 5x2+3x2 by x+1.

Solution

alt

The quotient is 5x2. The remainder is 0. We write the result as

5x2+3x2x+1=5x2

Analysis

This division problem had a remainder of 0. This tells us that (x+1) is a factor of 5x2+3x2, meaning 5x2+3x2=(x+1)(5x2), and that x=1 is a zero of p(x)=5x2+3x2.

Example 4.4.3

Divide 6x3+11x231x+15 by 3x2.

Solution

alt

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

6x3+11x231x+153x2=2x2+5x7+13x2

Analysis

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

(3x2)(2x2+5x7)+1=6x3+11x231x+15

Notice, as we write our result,

  • the dividend is 6x3+11x231x+15
  • the divisor is 3x2
  • the quotient is 2x2+5x7
  • the remainder is 1. Since the remainder is 1 we know that 3x2 is not a factor of 6x3+11x231x+15.

Synthetic Division

Since dividing by xc is a way to check if a number, c, is a zero of the polynomial, it would be nice to have a faster way to divide by xc than having to use long division every time. Happily, quicker ways have been discovered.

Let’s look back at the long division we did in Example 1 and try to streamline it. First, let’s distribute the negatives in the subtraction steps:

屏幕快照 2019-06-23 上午5.51.23.png

Next, observe that the terms x3, 6x2, and 7x are the exact opposite of the terms above them. Also note that the terms we ‘bring down’ (namely the 5x and 14) aren’t really necessary to recopy since we know they will cancel out when we subtract, so let's omit them, too:

屏幕快照 2019-06-23 上午5.51.55.png

Now, let’s move things up a bit and, for reasons which will become clear in a moment, copy the x3 into the last row.

屏幕快照 2019-06-23 上午5.52.19.png

Note that by arranging things in this manner, each term in the last row is obtained by adding the two terms above it. Notice also that the quotient polynomial can be obtained by dividing each of the first three terms in the last row by x and adding the results. If you take the time to work back through the original division problem, you will find that this is exactly the way we determined the quotient polynomial.

This means that we no longer need to write the quotient polynomial down, nor the x in the divisor, to determine our answer.

屏幕快照 2019-06-23 上午5.52.46.png

We’ve streamlined things quite a bit so far, but we can still do more. Let’s take a moment to remind ourselves where the 2x2, 12x and 14 came from in the second row. Each of these terms was obtained by multiplying the terms in the quotient, x2, 6x and 7, respectively, by the -2 in x2, then by -1 when we changed the subtraction to addition. Multiplying by -2 then by -1 is the same as multiplying by 2, so we replace the -2 in the divisor by 2. Furthermore, the coefficients of the quotient polynomial match the coefficients of the first three terms in the last row, so we now take the plunge and write only the coefficients of the terms to get

We have constructed a synthetic division tableau for this polynomial division problem. Let’s re-work our division problem using this tableau to see how it greatly streamlines the division process. To divide x3+4x25x14 by x2, we write 2 in the place of the divisor and the coefficients of x3+4x25x14 in for the dividend. Then "bring down" the first coefficient of the dividend.

屏幕快照 2019-06-23 上午5.55.20.png

Next, take the 2 from the divisor and multiply by the 1 that was "brought down" to get 2. Write this underneath the 4, then add to get 6.

屏幕快照 2019-06-23 上午5.58.18.png

Now take the 2 from the divisor times the 6 to get 12, and add it to the -5 to get 7.

屏幕快照 2019-06-23 上午5.59.04.png

Finally, take the 2 in the divisor times the 7 to get 14, and add it to the -14 to get 0.

屏幕快照 2019-06-23 上午5.59.43.png

The first three numbers in the last row of our tableau 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 x2+6x+7. The number in the box is the remainder. Synthetic division is our tool of choice for dividing polynomials by divisors of the form xc. It is important to note that it works only for these kinds of divisors. Also take note that when a polynomial (of degree at least 1) is divided by xc, the result will be a polynomial of exactly one less degree. Finally, it is worth the time to trace each step in synthetic division back to its corresponding step in long division.

Example 4.4.4

Use synthetic division to divide 5x32x2+1 by x3.

Solution

When setting up the synthetic division tableau, we need to enter 0 for the coefficient of x in the dividend. Doing so gives

屏幕快照 2019-06-23 上午6.01.07.png

Since the dividend was a third degree polynomial, the quotient is a quadratic polynomial with coefficients 5, 13 and 39. Our quotient is q(x)=5x2+13x+39 and the remainder is r(x)=118. We can express this as 5x32x2+1x3=5x2+13x+39+118x3

We also know that x3 is not a factor of 5x32x2+1 since the remainder is not zero.

Example 4.4.5

Divide x3+8 by x+2.

Solution

For this division, we rewrite x+2 as x(2) and proceed as before.

屏幕快照 2019-06-23 上午6.02.55.png

The quotient is x22x+4 and the remainder is zero. Since the remainder is zero, x+2 is a factor of x3+8.

x3+8=(x+2)(x22x+4)

We can write our answer as x3+8x+2=x22x+4

Note in the last example that there are placeholders for the missing x2 and x terms. When using synthetic division you must put place holders for the missing terms for the division to work properly.

You Try 4.4.1

Divide 4x48x25x by x3 using synthetic division.

Answer

屏幕快照 2019-06-23 上午6.06.29.png

4x48x25x divided by x3 is 4x3+12x2+28x+79 with remainder 237

Using this process allows us to find the real zeros of polynomials, presuming we can figure out at least one root. We’ll explore how to do that in the next section.

Example 4.4.6

The polynomial p(x)=4x44x311x2+12x3 has a horizontal intercept at x=12 with multiplicity 2. Find the other intercepts of p(x).

Solution

Since x=12 is an intercept with multiplicity 2, then x12 is a factor twice. Use synthetic division to divide by x12 twice.

屏幕快照 2019-06-23 上午6.04.51.png

From the first division, we get 4x44x311x2+12x3=(x12)(4x32x2x6) The second division tells us

4x44x311x2+12x3=(x12)(x12)(4x212)

To find the remaining intercepts, we set 4x212=0 and get x=±3.

Note this also means 4x44x311x2+12x3=4(x12)(x12)(x3)(x+3).

Let's summarize what we've learned about zeros and polynomial functions:

Connections Between Zeros, Factors, and Graphs of Polynomial Functions

Suppose p is a polynomial function of degree n1. The following statements are equivalent:

  • The real number c is a zero of p
  • p(c)=0
  • x=c is a solution to the polynomial equation p(x)=0
  • (xc) is a factor of p(x)
  • The point (c,0) is an x-intercept of the graph of y=p(x)

This page titled 4.4: Polynomial Division is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Katherine Skelton.

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