Difference between revisions of "Factor Theorem"
10000th User (talk | contribs) (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)'' | ||
+ | == Introduction == | ||
+ | == Theorem and Proof == | ||
+ | '''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>. | ||
− | + | *'''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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− | 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 == | ||
+ | |||
+ | == See also == | ||
+ | |||
+ | [[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 is a polynomial, then
is a factor
iff
.
- Proof: If
is a factor of
, then
, where
is a polynomial with
. Then
.
Now suppose that .
Apply division algorithm to get , where
is a polynomial with
and
is the remainder polynomial such that
.
This means that can be at most a constant polynomial.
Substitute and get
.
But is a constant polynomial and so
for all
.
Therefore, , which shows that
is a factor of
.