AoPSWiki

Try our innovative online adaptive learning system, Alcumus.
Over 1100 problems and 60+ video lessons. FREE!

Personal tools

Search

Toolbox

Factor Theorem

From AoPSWiki

Revision as of 06:22, 21 April 2008 by I like pie (Talk | contribs)

(diff) ← Older revision | Current revision (diff) | Newer revision → (diff)

The Factor Theorem says that if $P(x)$ is a polynomial, then $x-a$ is a factor of $P(x)$ iff $P(a)=0$ .

Proof

If $x - a$ is a factor of $P(x)$ , then $P(x) = (x - a)Q(x)$ , where $Q(x)$ is a polynomial with $\deg(Q(x)) = \deg(P(x)) - 1$ . Then $P(a) = (a - a)Q(a) = 0$ .

Now suppose that $P(a) = 0$ .

Apply division algorithm to get $P(x) = (x - a)Q(x) + R(x)$ , where $Q(x)$ is a polynomial with $\deg(Q(x)) = \deg(P(x)) - 1$ and $R(x)$ is the remainder polynomial such that $0\le\deg(R(x)) < \deg(x - a) = 1$ . This means that $R(x)$ can be at most a constant polynomial.

Substitute $x = a$ and get $P(a) = (a - a)Q(a) + R(a) = 0\Rightarrow R(a) = 0$ . Since $R(x)$ is a constant polynomial, $R(x) = 0$ for all $x$ .

Therefore, $P(x) = (x - a)Q(x)$ , which shows that $x - a$ is a factor of $P(x)$ .

This article is a stub. Help us out by expanding it.

Retrieved from "http://www.artofproblemsolving.com/Wiki/index.php/Factor_Theorem"

Categories: Stubs | Elementary algebra | Theorems

Looking for a challenging algebra text? Preparing for MATHCOUNTS or the AMC exams?
Check out Art of Problem Solving's Introduction to Algebra by Richard Rusczyk.

© Copyright 2008 AoPS Incorporated. All Rights Reserved. • Foundation • Privacy • Contact Us