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| − | ==Theorem==
| + | '''Remainder Theorem''' may refer to: |
| − | 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>.
| + | *[[Polynomial Remainder Theorem]] |
| − | | + | *[[Chinese Remainder Theorem]] |
| − | ==Proof==
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| − | 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
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| − | <cmath>p(x) = q(x) \cdot (x-a) + r(x)</cmath>
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| − | If <math>x = a</math>, then substituting for <math>x</math> results in
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| − | <cmath>p(a) = q(a) \cdot (a - a) + r(a)</cmath>
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| − | <cmath>p(a) = q(a) \cdot 0 + r(a)</cmath>
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| − | <math></math>p(a) = r(a)
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| − | ==Examples==
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| − | ===Introductory===
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| − | * What is the remainder when <math>x^2+2x+3</math> is divided by <math>x+1</math>?
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| − | ''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>.
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| − | * [[1961 AHSME Problems/Problem 22]] | |
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| − | {{stub}}
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Latest revision as of 16:42, 27 February 2022
Remainder Theorem may refer to:
