Difference between revisions of "Remainder Theorem"

Revision as of 11:47, 20 May 2018

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 $p(a) = q(a) \cdot (a - a) + r(a)$ $p(a) = q(a) \cdot 0 + r(a)$ $$ (Error compiling LaTeX. Unknown error_msg)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

This article is a stub. Help us out by expanding it.

@@ Line 1: / Line 1: @@
-=Theorem=
+==Theorem==
 The Remainder Theorem states that the remainder when the polynomial <math>P(x)</math> is divided by <math>x-a</math> (usually with synthetic division) is equal to the simplified value of <math>P(a)</math>.
-=Examples=
+==Proof==
-==Example 1==
+Let <math>\frac{p(x)}{x-a} = q(x) + \frac{r(x)}{x-a}</math>, where <math>p(x)</math> is the polynomial, <math>x-a</math> is the divisor, <math>q(x)</math> is the quotient, and <math>r(x)</math> is the remainder.  This equation can be rewritten as
-What is the remainder when <math>x^2+2x+3</math> is divided by <math>x+1</math>?
+<cmath>p(x) = q(x) \cdot (x-a) + r(x)</cmath>
+If <math>x = a</math>, then substituting for <math>x</math> results in
-==Solution==
+<cmath>p(a) = q(a) \cdot (a - a) + r(a)</cmath>
-Using synthetic or long division we obtain the quotient <math>1+\frac{2}{x^2+2x+3}</math>. In this case the remainder is <math>2</math>. However, we could've figured that out by evaluating <math>P(-1)</math>. Remember, we want the divisor in the form of <math>x-a</math>. <math>x+1=x-(-1)</math> so <math>a=-1</math>.
+<cmath>p(a) = q(a) \cdot 0 + r(a)</cmath>
+<math></math>p(a) = r(a)
+==Examples==
+===Introductory===
+* What is the remainder when <math>x^2+2x+3</math> is divided by <math>x+1</math>?
+''Solution'': Using synthetic or long division we obtain the quotient <math>1+\frac{2}{x^2+2x+3}</math>. In this case the remainder is <math>2</math>. However, we could've figured that out by evaluating <math>P(-1)</math>. Remember, we want the divisor in the form of <math>x-a</math>. <math>x+1=x-(-1)</math> so <math>a=-1</math>.
 <math>P(-1) = (-1)^2+2(-1)+3 = 1-2+3 = \boxed{2}</math>.
+* [[1961 AHSME Problems/Problem 22]]
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