Difference between revisions of "Polynomial remainder theorem"

Revision as of 16:27, 5 November 2021

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

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

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 that can be cracked by the remainder theorem or its adjacent ideas.

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 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 $2$ . $\square$ .

Example 2

[Insert problem involving the generalization of the remainder theorem]

@@ Line 2: / Line 2: @@
 == Proof ==
-We use the [[Euclidean polynomial division theorem]] with dividend <math>P(x)</math> and divisor <math>x-a</math>. The theorem 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>
+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>
 == Generalization ==

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