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| − | ==Theorem==
| + | #REDIRECT[[Polynomial remainder 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>.
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| − | ==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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| − | <cmath>p(a) = r(a)</cmath>
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| − | ==Extension==
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| − | An extension of the Remainder Theorem could be used to find the remainder of a polynomial when it is divided by a non-linear polynomial. Note that if <math>p(x)</math> is a polynomial, <math>q(x)</math> is the quotient, <math>d(x)</math> is a divisor, and <math>r(x)</math> is the remainder, the polynomial can be written as
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| − | <cmath>p(x) = q(x)d(x) + r(x)</cmath>
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| − | Note that the degree of <math>r(x)</math> is less than the degree of <math>d(x)</math>. Let <math>a_n</math> be a root of <math>d(x)</math>, where <math>n</math> is an integer and <math>1 \le n \le \text{deg } d</math>. That means for all <math>a_n</math>,
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| − | <cmath>p(a_n) = r(a_n)</cmath>
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| − | Thus, the points <math>(a_n,p(a_n))</math> are on the graph of the remainder. If all the roots of <math>d(x)</math> are unique, then a [[system of equations]] can be made to find the remainder <math>r(x)</math>.
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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>.
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| − | <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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| − | * [[1950 AHSME Problems/Problem 20]]
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| − | ===Intermediate===
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| − | * [[1969 AHSME Problems/Problem 34]]
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| − | [[Category:Algebra]]
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| − | [[Category:Polynomials]]
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| − | [[Category:Theorems]]
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