Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "Factor Theorem"

(Found a proof I provided a long time ago, a gem hidden in the forums)
 
Line 1: Line 1:
 
{{stub}}
 
{{stub}}
 
{{wikify}}
 
{{wikify}}
 +
''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>.
  
−
== Headline text ==
+
*'''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>.
−
 
 
−
== Headline text ==
 
−
 
 
−
== Headline text ==
 
−
 
 
−
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>.
Line 22: Line 20:
  
 
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>.
 +
 +
 +
== Problems ==
 +
 +
== See also ==
 +
 +
[[Category: Polynomials]]

Revision as of 13:44, 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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=Factor_Theorem&oldid=19528"