Difference between revisions of "Factor Theorem"

Revision as of 19:43, 28 March 2022

The Factor Theorem says that if $P(x)$ is a polynomial, then $\displaystyle{x-a}$ 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 $P(x)$ .

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Revision as of 13:06, 14 July 2021 (view source) Etmetalakret (talk \| contribs) ← Older edit		Revision as of 19:43, 28 March 2022 (view source) Andrew2019 (talk \| contribs) Newer edit →
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−	The '''Factor Theorem''' says that if <math>P(x)</math> is a [[polynomial]], then <math>x-a</math> is a [[factor]] of <math>P(x)</math> if <math>P(a)=0</math>.	+	The '''Factor Theorem''' says that if <math>P(x)</math> is a [[polynomial]], then <math>\displaystyle{x-a}</math> is a [[factor]] of <math>P(x)</math> if <math>P(a)=0</math>.

	==Proof==		==Proof==

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Difference between revisions of "Factor Theorem"

Revision as of 19:43, 28 March 2022

Proof