﻿ the Remainder Theorem

# THE REMAINDER THEOREM

• PRACTICE (online exercises and printable worksheets)

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 $\,P(4)\,$.

SOLUTION:

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

The remainder is $\,1293\,$, so $\,P(4) = 1293\,$.
How easy is that!?

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

\displaystyle \begin{align} P(4) &= 2\cdot4^5 - 3\cdot4^4 + 5\cdot 4 - 7\cr &= 2\cdot1024 - 3\cdot256 + 20 - 7\cr &= 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:

By the division algorithm, there exist unique polynomials $\,Q(x)\,$ (the quotient) and $\,R(x)\,$ (the remainder) such that $$P(x) = Q(x)(x-c) + R(x)\,,$$ 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, $\,R\,$: $$P(x) = Q(x)(x-c) + R$$ Evaluating $\,P\,$ at $\,c\,$ gives: $$P(c) \quad = \quad Q(c)(c - c) + R \quad = \quad Q(c)\cdot 0 + R \quad = \quad 0 + R \quad = \quad R$$
QED

Master the ideas from this section