Difference between revisions of "Polynomial Remainder Theorem"

Latest revision as of 23:39, 1 April 2023

In algebra, the Polynomial Remainder Theorem states that the remainder upon dividing any polynomial $P(x)$ by a linear polynomial $x-a$ , both with complex coefficients, is equal to $P(a)$ .

Proof

By polynomial division with dividend $P(x)$ and divisor $x-a$ , that exist a quotient $Q(x)$ and remainder $R(x)$ such that $P(x) = (x-a) Q(x) + R(x)$ with $\deg R(x) < \deg (x-a)$ . We wish to show that $R(x)$ is equal to the constant $P(a)$ . Because $\deg (x-a) = 1$ , $\deg R(x) < 1$ . Therefore, $\deg R(x) = 0$ , and so the $R(x)$ is a constant.

Let this constant be $r$ . We may substitute this into our original equation and rearrange to yield $r = P(x) - (x-a) Q(x).$ When $x = a$ , this equation becomes $r = P(a)$ . Hence, the remainder upon diving $P(x)$ by $x-a$ is equal to $P(a)$ . $\square$

Generalization

The strategy used in the above proof can be generalized to divisors with degree greater than $1$ . A more general method, with any dividend $P(x)$ and divisor $D(x)$ , is to write $R(x) = D(x) Q(x) - P(x)$ , and then substitute the zeroes of $D(x)$ to eliminate $Q(x)$ and find values of $R(x)$ . 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 $x^2+2x+3$ is divided by $x+1$ ?

Solution: Although one could use long or synthetic division, the Polynomial Remainder Theorem provides a significantly shorter solution. Note that $P(x) = x^2 + 2x + 3$ , and $x-a = x+1$ . A common mistake is to forget to flip the negative sign and assume $a = 1$ , but simplifying the linear equation yields $a = -1$ . Thus, the answer is $P(-1)$ , or $(-1)^2 + 2(-1) + 3$ , which is equal to $2$ . $\square$ .

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 == Proof ==
-We use Euclidean polynomial division with dividend <math>P(x)</math> and divisor <math>x-a</math>. The result states that there exists 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>f(a)</math>. Because <math>\deg (x-a) = 1</math>, <math>\deg R(x) < 1</math>. Hence, <math>R(x)</math> is a constant, <math>r</math>. Plugging this into our original equation and rearranging a bit yields <cmath>r = P(x) - (x-a) Q(x).</cmath> After substituting <math>x=a</math> into this equation, we deduce that <math>r = P(a)</math>; thus, the remainder upon diving <math>P(x)</math> by <math>x-a</math> is equal to <math>P(a)</math>, as desired. <math>\square</math>
+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.
+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 ==

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Difference between revisions of "Polynomial Remainder Theorem"

Latest revision as of 23:39, 1 April 2023

Contents

Proof

Generalization

Examples

Example 1

More examples

See also