Department of Mathematics --- College of Science --- University of Utah

Mathematics 1010 online

Working with Polynomials

Polynomials occur ubiquitously in applications. A number of techniques have been developed for working with them and exploiting their special structure. In this class you will learn how to evaluate, factor, and divide polynomials.

Evaluating a polynomial

Consider a polynomial such as

$\displaystyle p(x) = x^3 - 6x^2 + 11x -
6. \qquad(*) $

It is clear, for example, that $ p(1) = 1-6+11-6=0. $ Suppose you want to know $ p(7) $. You could evaluate those powers of $ 7 $ in $ (*) $, multiply with the coefficients, and add the individual terms to get

p(7) &=& 7^3 -6 \times 7^2 + 11 \times 7 ...
...7 -6 \\
&=& 343 - 294 + 77 -6\\
&=& 120.\\
\end{array} \end{displaymath}

It turns out that this is a clumsy procedure. The evaluation becomes much simpler if we rewrite $ p $ as

$\displaystyle p(x) = \big((x-6)x+11\big)x-6\qquad (**) $

(You should take a moment to convince yourself that $ (*) $ and $ (**) $ are in fact equivalent.) Evaluating $ p $ in the form $ (**) $ is much easier since we don't have to compute powers. In our special case we obtain:

p(7) &=& \big((7-6)\times 7+11\big)\times...
...s 7 - 6 \\
&=& 18 \times 7 - 6 \\
&=& 120.\\
\end{array} \end{displaymath}

We get the same answer, of course, but the computation is simpler and it involves fewer basic operations. The advantages of this form are more pronounced for polynomials of high degree. The above technique works in general and is described in the literature as synthetic division (for reasons discussed below), nested multiplication (for reasons that are obvious when considering $ (**) $), or Horner's Scheme (after the English Algebraist William George Horner, 1786-1837). It's a good rule of thumb that if something is known under several names it is usually powerful or otherwise important.

You can use synthetic division to evaluate a polynomial in your head, or using a calculator (storing just one number, the value of $ x $). The usual way to do it on paper is to construct a table which has the coefficients of the polynomial in the first row, and the intermediate results of the Calculation in the second and third rows:

1 & -6 & 11 & - 6 \\
& 7 &7 & 126\\
1 & 1 & 18& 120 \\
\end{array} \end{displaymath}

Each entry in the third row is the sum of the entries above it. In the first column there is a blank in the second row that you can think of as zero. Each entry in the second row is obtained by multiplying the entry to the lower left of it with the number at which we evaluate the polynomial (in this case 7). Usually we take note of that number by writing it to the left of the array in the first row:

7& 1 & -6 & 11 & - 6 \\
& & 7 &7 & 126\\
& 1 & 1 & 18& 120 \\
\end{array} \end{displaymath}

The final result of the evaluation (in this case $ p(7)=120 $) is given (and underlined) in the lower right corner of the array.

The process is called synthetic division because it can be thought of as dividing our polynomial by $ x-7 $ with remainder. The quotient is a polynomial whose coefficients are the other entries in the last row. In this case,

$\displaystyle p(x) = x^3 - 6x^2 + 11x - 6 = (x^2 + x + 18)(x-7) +
120. $

It's a little involved but a good exercise to see that this works in general. Send me an e-mail or talk with me if you are interested in more information.

Factoring a Polynomial

We write a polynomial in standard form as a sum of monomials. In factored form a polynomial is written as a product of lower degree polynomials. It is completely factored if all factors are of as low a degree as possible.

For example, you can easily check that our polynomial $ p $ satisfies

p(x) &= & x^3 - 6x^2 + 11x - 6\\
& = &...
...x+6)) \\
& = & (x-1)(x-2)(x-3) \\

The significance of having a polynomial $ p $ in factored form is that it makes it easier to solve equations of the form $ p(x)=0 $. A solution of that equation is called a root or zero of $ p $. A product is zero if and only if one of the factors is zero, so to find the roots of $ p $ we need to look only at the roots of the individual factors. It is clear from $ (***) $ that the roots of our particular polynomial $ p $ are $ 1 $, $ 2 $, and $ 3 $. This is not at all obvious by looking at the original definition $ (*) $ of $ p $.

Factoring is emphasized in a class like this as a means of solving polynomial equations. It works beautifully if you are able to find the factors. However, the usual flow is often the other way, if you really want factors you find them by first finding the roots of a polynomial.

There is, however, one special case, where you find factors that are common to two (or more) polynomials, usually for the purpose of canceling them in a ratio of polynomials. That subject deserves its own page on long division and the Euclidean Algorithm .