Difference between revisions of "Factor Theorem"

Revision as of 13:59, 18 July 2023

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

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 Remainder Theorem 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.

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 Apply [[Remainder Theorem]] to get <math>P(x) = (x - a)Q(x) + R(x)</math>, where <math>Q(x)</math> is a polynomial with <math>\deg(Q(x)) = \deg(P(x)) - 1</math> and <math>R(x)</math> is the remainder polynomial such that <math>0\le\deg(R(x)) < \deg(x - a) = 1</math>. This means that <math>R(x)</math> can be at most a [[constant]] polynomial.
-Substitute <math>x = a</math> and get <math>P(a) = (a - a)Q(a) + R(a) = 0\Rightarrow R(a) = 0</math>. Since <math>R(x)</math> is a constant polynomial, <math>R(x) = 0</math> for all <math>x</math>.
+Substitute <math>x = a</math> and get <math>P(a) = (a - a)Q(a) + R(a) = 0\Rightarrow R(a) = 0</math>. Since <math>R(x)</math> is a  constant polynomial, <math>R(x) = 0</math> for all <math>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>.
 {{stub}}
-m
 [[Category:Algebra]]
 [[Category:Polynomials]]
 [[Category:Theorems]]

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Difference between revisions of "Factor Theorem"

Revision as of 13:59, 18 July 2023

Proof