Difference between revisions of "Factor Theorem"
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Now suppose that <math>P(a) = 0</math>. | Now suppose that <math>P(a) = 0</math>. | ||
− | 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 | + | 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>. |
Revision as of 22:52, 4 July 2021
The Factor Theorem says that if is a polynomial, then
is a factor of
if
.
Proof
If is a factor of
, then
, where
is a polynomial with
. Then
.
Now suppose that .
Apply Remainder Theorem 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
. Since
is a constant polynomial,
for all
.
Therefore, , which shows that
is a factor of
.
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