3.4: Polynomial Division (2024)

Long Division of Polynomials

Now, let's revisit the the polynomial \(f(x) = x^3 + 4x^2-5x-14\). 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 \(x^{3} +4x^{2} -5x-14=(x-2)\) (something). To find that "something," we can use polynomial division.

Now, if we want to find the other x-intercepts, we now solve \(x^2+6x+7=0\). Since this doesn't factor nicely we can complete the square or use the Quadratic Formula to get \(x = -3 \pm \sqrt{2}\). 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.

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 \(x-c\), then the remainder will be zero or a constant.

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

If \(x-c\) is a factor of the polynomial \(p\), then \(p(x)=(x-c)q(x)\) for some polynomial \(q\). Then \(p(c)=(c-c)q(c)=0\), 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 \(x-c\), then the remainder will be zero, which means \(x-c\) 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.

Synthetic Division

Since dividing by \(x-c\) 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 \(x-c\) 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:

Next, observe that the terms \(-x^{3}\), \(-6x^{2}\), and \(-7x\) are the exact opposite of the terms above them. Also note that the terms we ‘bring down’ (namely the \(\mathrm{-}\)5x and \(\mathrm{-}\)14) aren’t really necessary to recopy since we know they will cancel out when we subtract, so let's omit them, too:

Now, let’s move things up a bit and, for reasons which will become clear in a moment, copy the \(x^{3}\) into the last row.

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.

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 \(2x^{2}\), \(12x\) and 14 came from in the second row. Each of these terms was obtained by multiplying the terms in the quotient, \(x^{2}\), 6x and 7, respectively, by the -2 in \(x - 2\), 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 \(x^{3} +4x^{2} -5x-14\) by \(x-2\), we write 2 in the place of the divisor and the coefficients of \(x^{3} +4x^{2} -5x-14\) in for the dividend. Then "bring down" the first coefficient of the dividend.

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.

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

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

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 \(x^{2} +6x+7\). The number in the box is the remainder. Synthetic division is our tool of choice for dividing polynomials by divisors of the form \(x - c\). 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 \(x - c\), 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.

Note in the last example that there are placeholders for the missing \(x^2\) and \(x\) terms. When using synthetic division you must put place holders for the missing terms for the division to work properly.

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.

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