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| − | In [[algebra]], the '''polynomial remainder theorem''' states that the remainder upon [[Synthetic division | dividing]] any [[polynomial]] <math>P(x)</math> by a linear polynomial <math>x-a</math>, both with [[Complex number | complex]] coefficients, is equal to <math>P(a)</math>.
| + | #REDIRECT[[Polynomial Remainder Theorem]] |
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| − | == Proof ==
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| − | We use Euclidean polynomial division with dividend <math>P(x)</math> and divisor <math>x-a</math>. The result states that there exists a quotient <math>Q(x)</math> and remainder <math>R(x)</math> such that <cmath>P(x) = (x-a) Q(x) + R(x),</cmath> with <math>\deg R(x) < \deg (x-a)</math>. We wish to show that <math>R(x)</math> is equal to the constant <math>f(a)</math>. Because <math>\deg (x-a) = 1</math>, <math>\deg R(x) < 1</math>. Hence, <math>R(x)</math> is a constant, <math>r</math>. Plugging this into our original equation and rearranging a bit yields <cmath>r = P(x) - (x-a) Q(x).</cmath> After substituting <math>x=a</math> into this equation, we deduce that <math>r = P(a)</math>; thus, the remainder upon diving <math>P(x)</math> by <math>x-a</math> is equal to <math>P(a)</math>, as desired. <math>\square</math>
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| − | == Generalization ==
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| − | The strategy used in the above proof can be generalized to divisors with degree greater than <math>1</math>. A more general method, with any dividend <math>P(x)</math> and divisor <math>D(x)</math>, is to write <math>R(x) = D(x) Q(x) - P(x)</math>, and then substitute the zeroes of <math>D(x)</math> to eliminate <math>Q(x)</math> and find values of <math>R(x)</math>. Example 2 showcases this strategy.
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| − | == Examples==
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| − | Here are some problems that can be cracked by the remainder theorem or its adjacent ideas.
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| − | === Example 1 ===
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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''': Although one could use long or synthetic division, the remainder theorem provides a significantly shorter solution. Note that <math>P(x) = x^2 + 2x + 3</math>, and <math>x-a = x+1</math>. A common mistake is to forget to flip the negative sign and assume <math>a = 1</math>, but simplifying the linear equation yields <math>a = -1</math>. Thus, the answer is <math>P(-1)</math>, or <math>(-1)^2 + 2(-1) + 3</math>, which is equal to <math>2</math>. <math>\square</math>.
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| − | === Example 2 ===
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| − | [Insert problem involving the generalization of the remainder theorem]
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| − | === More examples ===
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| − | * [[1950 AHSME Problems/Problem 20 | 1950 ASHME Problem 20]]
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| − | * [[1961 AHSME Problems/Problem 22 | 1961 ASHME Problem 22]]
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| − | * [[1969 AHSME Problems/Problem 34 | 1969 ASHME Problem 34]]
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| − | == See also ==
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| − | * [[Polynomial]]
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| − | * [[Euclidean polynomial division theorem]]
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| − | * [[Factor theorem]]
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| − | [[Category:Algebra]] [[Category:Polynomials]] [[Category:Theorems]]
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