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Difference between revisions of "Polynomial Remainder Theorem"
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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>.
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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>.
  
 
== Proof ==
 
== 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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By polynomial division with dividend <math>P(x)</math> and divisor <math>x-a</math>, that exist 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>P(a)</math>. Because <math>\deg (x-a) = 1</math>, <math>\deg R(x) < 1</math>. Therefore, <math>\deg R(x) = 0</math>, and so the <math>R(x)</math> is a constant.
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Let this constant be <math>r</math>. We may substitute this into our original equation and rearrange to yield <cmath>r = P(x) - (x-a) Q(x).</cmath> When <math>x = a</math>, this equation becomes <math>r = P(a)</math>. Hence, the remainder upon diving <math>P(x)</math> by <math>x-a</math> is equal to <math>P(a)</math>. <math>\square</math>
  
 
== Generalization ==
 
== Generalization ==
 
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.
 
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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== Problems==
 
Here are some problems with solutions that utilize the Polynomial Remainder Theorem and its generalization.
 
Here are some problems with solutions that utilize the Polynomial Remainder Theorem and its generalization.
  
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'''Solution''': Although one could use long or synthetic division, the Polynomial 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>.  
 
'''Solution''': Although one could use long or synthetic division, the Polynomial 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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=== More examples ===
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=== More Problems ===
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* [[1950 AHSME Problems/Problem 20 | 1950 ASHME Problem 20]]
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* [[1950 AHSME Problems/Problem 20|1950 AHSME Problem 20]]
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* [[1961 AHSME Problems/Problem 22 | 1961 ASHME Problem 22]]
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* [[1961 AHSME Problems/Problem 22|1961 AHSME Problem 22]]
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* [[1969 AHSME Problems/Problem 34 | 1969 ASHME Problem 34]]
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* [[1969 AHSME Problems/Problem 34|1969 AHSME Problem 34]]
  
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== See also ==
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== See Also ==
 
* [[Polynomial]]
 
* [[Polynomial]]
 
* [[Factor theorem]]
 
* [[Factor theorem]]
  
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[[Category:Algebra]] [[Category:Polynomials]] [[Category:Theorems]]
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[[Category:Algebra]]
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[[Category:Polynomials]]
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[[Category:Theorems]]
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{{stub}}

Latest revision as of 13:00, 11 March 2025

In algebra, the Polynomial Remainder Theorem states that the remainder upon dividing any polynomial $P(x)$ by a linear polynomial $x-a$, both with complex coefficients, is equal to $P(a)$.

Proof

By polynomial division with dividend $P(x)$ and divisor $x-a$, that exist a quotient $Q(x)$ and remainder $R(x)$ such that \[P(x) = (x-a) Q(x) + R(x)\] with $\deg R(x) < \deg (x-a)$. We wish to show that $R(x)$ is equal to the constant $P(a)$. Because $\deg (x-a) = 1$, $\deg R(x) < 1$. Therefore, $\deg R(x) = 0$, and so the $R(x)$ is a constant.

Let this constant be $r$. We may substitute this into our original equation and rearrange to yield \[r = P(x) - (x-a) Q(x).\] When $x = a$, this equation becomes $r = P(a)$. Hence, the remainder upon diving $P(x)$ by $x-a$ is equal to $P(a)$. $\square$

Generalization

The strategy used in the above proof can be generalized to divisors with degree greater than $1$. A more general method, with any dividend $P(x)$ and divisor $D(x)$, is to write $R(x) = D(x) Q(x) - P(x)$, and then substitute the zeroes of $D(x)$ to eliminate $Q(x)$ and find values of $R(x)$. Example 2 showcases this strategy.

Problems

Here are some problems with solutions that utilize the Polynomial Remainder Theorem and its generalization.

Example 1

What is the remainder when $x^2+2x+3$ is divided by $x+1$?

Solution: Although one could use long or synthetic division, the Polynomial Remainder Theorem provides a significantly shorter solution. Note that $P(x) = x^2 + 2x + 3$, and $x-a = x+1$. A common mistake is to forget to flip the negative sign and assume $a = 1$, but simplifying the linear equation yields $a = -1$. Thus, the answer is $P(-1)$, or $(-1)^2 + 2(-1) + 3$, which is equal to $2$. $\square$.

More Problems

See Also

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