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 $p(a) = q(a) \cdot (a - a) + r(a)$ $p(a) = q(a) \cdot 0 + r(a)$ $p(a) = r(a)$

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 $a=-1$ . $P(-1) = (-1)^2+2(-1)+3 = 1-2+3 = \boxed{2}$ .

1961 AHSME Problems/Problem 22

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@@ Line 8: / Line 8: @@
 <cmath>p(a) = q(a) \cdot (a - a) + r(a)</cmath>
 <cmath>p(a) = q(a) \cdot 0 + r(a)</cmath>
-<math></math>p(a) = r(a)
+<cmath>p(a) = r(a)</cmath>
 ==Examples==

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