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Difference between revisions of "Factor Theorem"

(Found a proof I provided a long time ago, a gem hidden in the forums)
 
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''This page is under heavy construction--[[User:10000th User|10000th User]] 13:44, 15 November 2007 (EST)''
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== Introduction ==
  
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== Theorem and Proof ==
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'''Theorem:''' If <math>P(x)</math> is a polynomial, then <math>x-a</math> is a factor <math>P(x)</math> iff <math>P(a)=0</math>.
  
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== Headline text ==
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*'''Proof:''' If <math>x - a</math> is a factor of <math>P(x)</math>, then <math>P(x) = (x - a)Q(x)</math>, where <math>Q(x)</math> is a polynomial with <math>\deg(Q(x)) = \deg(P(x)) - 1</math>. Then <math>P(a) = (a - a)Q(a) = 0</math>.
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== Headline text ==
 
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If <math>x - a</math> is a factor of <math>P(x)</math>, then <math>P(x) = (x - a)Q(x)</math>, where <math>Q(x)</math> is a polynomial with <math>\deg(Q(x)) = \deg(P(x)) - 1</math>. Then <math>P(a) = (a - a)Q(a) = 0</math>.
 
  
 
Now suppose that <math>P(a) = 0</math>.
 
Now suppose that <math>P(a) = 0</math>.
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Therefore, <math>P(x) = (x - a)Q(x)</math>, which shows that <math>x - a</math> is a factor of <math>P(x)</math>.
 
Therefore, <math>P(x) = (x - a)Q(x)</math>, which shows that <math>x - a</math> is a factor of <math>P(x)</math>.
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== Problems ==
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== See also ==
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[[Category: Polynomials]]

Revision as of 14:44, 15 November 2007

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

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