Difference between revisions of "Factor Theorem"
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==Theorem== | ==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>. | + | 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== | ==Proof== | ||
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Now suppose that <math>P(a) = 0</math>. | Now suppose that <math>P(a) = 0</math>. | ||
− | Apply division algorithm 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>. | + | Apply division [[algorithm]] 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. | + | 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>. | Substitute <math>x = a</math> and get <math>P(a) = (a - a)Q(a) + R(a) = 0\Rightarrow R(a) = 0</math>. | ||
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== See also == | == See also == | ||
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{{stub}} | {{stub}} |
Revision as of 15:28, 9 February 2008
The Factor Theorem is a theorem relating to polynomials
Contents
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
.
Problems
See also
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