Difference between revisions of "Polynomial Remainder Theorem"
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Etmetalakret (talk | contribs) m (This article is still TERRIBLE but the proof makes a little more sense.) |
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== Proof == | == Proof == | ||
− | + | By polynomial division with dividend <math>P(x)</math> and divisor <math>x-a</math>, that exist a quotient <math>Q(x)</math> and remainder <math>R(x)</math> such that <cmath>P(x) = (x-a) Q(x) + R(x)</cmath> with <math>\deg R(x) < \deg (x-a)</math>. We wish to show that <math>R(x)</math> is equal to the constant <math>P(a)</math>. Because <math>\deg (x-a) = 1</math>, <math>\deg R(x) < 1</math>. Therefore, <math>\deg R(x) = 0</math>, and so the <math>R(x)</math> is a constant. | |
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+ | Let this constant be <math>r</math>. We may substitute this into our original equation and rearrange to yield <cmath>r = P(x) - (x-a) Q(x).</cmath> When <math>x = a</math>, this equation becomes <math>r = P(a)</math>. Hence, the remainder upon diving <math>P(x)</math> by <math>x-a</math> is equal to <math>P(a)</math>. <math>\square</math> | ||
== Generalization == | == Generalization == | ||
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== Examples== | == Examples== | ||
− | Here are some problems with solutions that utilize the | + | Here are some problems with solutions that utilize the Polynomial Remainder Theorem and its generalization. |
=== Example 1 === | === Example 1 === | ||
''What is the remainder when <math>x^2+2x+3</math> is divided by <math>x+1</math>?'' | ''What is the remainder when <math>x^2+2x+3</math> is divided by <math>x+1</math>?'' | ||
− | '''Solution''': Although one could use long or synthetic division, the | + | '''Solution''': Although one could use long or synthetic division, the Polynomial Remainder Theorem provides a significantly shorter solution. Note that <math>P(x) = x^2 + 2x + 3</math>, and <math>x-a = x+1</math>. A common mistake is to forget to flip the negative sign and assume <math>a = 1</math>, but simplifying the linear equation yields <math>a = -1</math>. Thus, the answer is <math>P(-1)</math>, or <math>(-1)^2 + 2(-1) + 3</math>, which is equal to <math>2</math>. <math>\square</math>. |
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=== More examples === | === More examples === |
Latest revision as of 23:39, 1 April 2023
In algebra, the Polynomial Remainder Theorem states that the remainder upon dividing any polynomial by a linear polynomial
, both with complex coefficients, is equal to
.
Proof
By polynomial division with dividend and divisor
, that exist a quotient
and remainder
such that
with
. We wish to show that
is equal to the constant
. Because
,
. Therefore,
, and so the
is a constant.
Let this constant be . We may substitute this into our original equation and rearrange to yield
When
, this equation becomes
. Hence, the remainder upon diving
by
is equal to
.
Generalization
The strategy used in the above proof can be generalized to divisors with degree greater than . A more general method, with any dividend
and divisor
, is to write
, and then substitute the zeroes of
to eliminate
and find values of
. Example 2 showcases this strategy.
Examples
Here are some problems with solutions that utilize the Polynomial Remainder Theorem and its generalization.
Example 1
What is the remainder when is divided by
?
Solution: Although one could use long or synthetic division, the Polynomial Remainder Theorem provides a significantly shorter solution. Note that , and
. A common mistake is to forget to flip the negative sign and assume
, but simplifying the linear equation yields
. Thus, the answer is
, or
, which is equal to
.
.