Factor Theorem

Revision as of 13:44, 15 November 2007 by 10000th User (talk | contribs)

This article is a stub. Help us out by expanding it. Template:Wikify This page is under heavy construction--10000th User 13:44, 15 November 2007 (EST)

Introduction

Theorem and Proof

Theorem: If $P(x)$ is a polynomial, then $x-a$ is a factor $P(x)$ 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 $P(a) = (a - a)Q(a) + R(a) = 0\Rightarrow R(a) = 0$.

But $R(x)$ is a constant polynomial and so $R(x) = 0$ for all $x$.

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


Problems

See also