﻿ The Remainder Theorem

# The Remainder Theorem

This section gives a useful application of synthetic division and the division algorithm, called the Remainder Theorem or the Polynomial Remainder Theorem.

The Remainder Theorem provides an efficient way to evaluate a polynomial at a given number: that is, to find $\,P(c)\,$ for a given polynomial $\,P(x)\,$ and a given real number $\,c\,.$

How?  Just divide $\,P(x)\,$ by $\,x - c\,$ and then take the remainder! After the example below is a precise statement of the Remainder Theorem, and why it works.

## Example

Let $\,P(x) = 2x^5 - 3x^4 + 5x - 7\,.$   Find $\,P(\color{red}{4})\,.$

Solution: Divide $\,P(x)\,$ by $\,x-\color{red}{4}\,,$ using synthetic division. Since we're dividing by $\,x - \color{red}{4}\,,$ the number $\,\color{red}{4}\,$ goes in the little box:

 $\color{red}{4}$ $2$ $-3$ $0$ $0$ $5$ $-7$ $8$ $20$ $80$ $320$ $1300$ $2$ $5$ $20$ $80$ $325$ $\color{blue}{1293}$

The remainder is $\,\color{blue}{1293}\,,$ so $\,P(\color{red}{4}) = \color{blue}{1293}\,.$ How easy is that!?

Let's check this result with old-fashioned function evaluation:

\begin{align} \cssId{s17}{P(4)\ } &\cssId{s18}{= 2\cdot4^5 - 3\cdot4^4 + 5\cdot 4 - 7}\cr &\cssId{s19}{= 2\cdot1024 - 3\cdot256 + 20 - 7}\cr &\cssId{s20}{= 1293} \end{align}

Notice that function evaluation requires powers (like $\,4^5\,$), whereas the Remainder Theorem requires only multiplication and addition.

The Remainder Theorem finding $\,P(c)\,$ for a polynomial $\,P\,$

Let $\,P(x)\,$ be a polynomial, and let $\,c\,$ be a real number.

Then, $\,P(c)\,$ is the remainder when $\,P(x)\,$ is divided by $\,x - c\,.$

Proof:  there exist unique polynomials $\,Q(x)\,$ (the quotient) and $\,R(x)\,$ (the remainder) such that

$$\cssId{s30}{P(x) = Q(x)(x-c) + R(x)}\,,$$

where either

• $\,R(x) = 0\,,$ or
• the degree of $\,R(x)\,$ is less than the degree of $\,x - c\,$ (which is $\,1\,$)

If $\,R(x)\,$ is zero, then it's a number—a constant.

If $\,R(x)\,$ has degree zero (which is the only degree less than $\,1\,$), then it is a constant.

So, in both cases, $\,R(x)\,$ is a constant—it doesn't depend on $\,x\,$ at all. To reflect this, give $\,R(x)\,$ a simpler name, $\,R\,$:

$$\cssId{s38}{P(x) = Q(x)(x-c) + R}$$

Evaluating $\,P\,$ at $\,c\,$ gives:

\begin{align} \cssId{s40}{P(c)\ } &\cssId{s41}{= Q(c)(c - c) + R}\cr\cr &\cssId{s42}{= Q(c)\cdot 0 + R}\cr\cr &\cssId{s43}{= 0 + R}\cr\cr &\cssId{s44}{= R} \end{align}