Difference between revisions of "Remainder Theorem"

(Theorem)
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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>
 
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>
  
==What it's not==
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=Examples=
This is not the same as the [[Chinese Remainder Theorem]]
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==Example 1==
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What is the remainder in <math>\frac{x^2+2x+3}{x+1}</math>?
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==Solution==
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Using synthetic or long division we obtain the quotient <math>x+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>x=-1</math>.
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<math>P(-1) = (-1)^2+2(-1)+3 = 1-2+3 = \boxed{2}</math>

Revision as of 11:49, 20 November 2011

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 $P(a)$

Examples

Example 1

What is the remainder in $\frac{x^2+2x+3}{x+1}$?

Solution

Using synthetic or long division we obtain the quotient $x+1+\frac{2}{x^2+2x+3}$. 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 $x=-1$.

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