Difference between revisions of "Remainder Theorem"

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Theorem

The Remainder Theorem states that the remainder when the polynomial $P(x)$ is divided by $x-a$ (usually with synthetic division) is equal to the simplified value of $P(a)$ .

Proof

Let $\frac{p(x)}{x-a} = q(x) + \frac{r(x)}{x-a}$ , where $p(x)$ is the polynomial, $x-a$ is the divisor, $q(x)$ is the quotient, and $r(x)$ is the remainder. This equation can be rewritten as $p(x) = q(x) \cdot (x-a) + r(x)$ If $x = a$ , then substituting for $x$ results in $p(a) = q(a) \cdot (a - a) + r(a)$ $p(a) = q(a) \cdot 0 + r(a)$ $p(a) = r(a)$

Extension

An extension of the Remainder Theorem could be used to find the remainder of a polynomial when it is divided by a non-linear polynomial. Note that if $p(x)$ is a polynomial, $q(x)$ is the quotient, $d(x)$ is a divisor, and $r(x)$ is the remainder, the polynomial can be written as $p(x) = q(x)d(x) + r(x)$ Note that the degree of $r(x)$ is less than the degree of $d(x)$ . Let $a_n$ be a root of $d(x)$ , where $n$ is an integer and $1 \le n \le \text{deg } d$ . That means for all $a_n$ , $p(a_n) = r(a_n)$ Thus, the points $(a_n,p(a_n))$ are on the graph of the remainder. If all the roots of $d(x)$ are unique, then a system of equations can be made to find the remainder $r(x)$ .

Examples

Introductory

What is the remainder when $x^2+2x+3$ is divided by $x+1$ ?

Solution: Using synthetic or long division we obtain the quotient $1+\frac{2}{x^2+2x+3}$ . In this case the remainder is $2$ . However, we could've figured that out by evaluating $P(-1)$ . Remember, we want the divisor in the form of $x-a$ . $x+1=x-(-1)$ so $a=-1$ . $P(-1) = (-1)^2+2(-1)+3 = 1-2+3 = \boxed{2}$ .