Difference between revisions of "Remainder Theorem"

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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>.
 
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>
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<math>P(-1) = (-1)^2+2(-1)+3 = 1-2+3 = \boxed{2}</math>.
  
 
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Revision as of 16:55, 2 July 2017

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 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 $a=-1$.

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

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