Difference between revisions of "Remainder Theorem"

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=Theorem=
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==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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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>.
  
=Examples=
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==Proof==
==Example 1==
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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
What is thé remainder in <math>\frac{x^2+2x+3}{x+1}</math>?
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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>
  
==Solution==
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==Extension==
Using synthetic or long division we obtain the quotient <math>x+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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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>.
  
<math>P(-1) = (-1)^2+2(-1)+3 = 1-2+3 = \boxed{2}</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]]
  
{{stub}}
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===Intermediate===
hello
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* [[1969 AHSME Problems/Problem 34]]
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[[Category:Polynomials]]
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[[Category:Algebra]]

Revision as of 13:12, 4 December 2020

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 $P(a)$.

Proof

Let $\frac{p(x)}{x-a} = q(x) + \frac{r(x)}{x-a}$, where $p(x)$ is the polynomial, $x-a$ is the divisor, $q(x)$ is the quotient, and $r(x)$ is the remainder. This equation can be rewritten as \[p(x) = q(x) \cdot (x-a) + r(x)\] If $x = a$, then substituting for $x$ results in \[p(a) = q(a) \cdot (a - a) + r(a)\] \[p(a) = q(a) \cdot 0 + r(a)\] \[p(a) = r(a)\]

Extension

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 $p(x)$ is a polynomial, $q(x)$ is the quotient, $d(x)$ is a divisor, and $r(x)$ is the remainder, the polynomial can be written as \[p(x) = q(x)d(x) + r(x)\] Note that the degree of $r(x)$ is less than the degree of $d(x)$. Let $a_n$ be a root of $d(x)$, where $n$ is an integer and $1 \le n \le \text{deg } d$. That means for all $a_n$, \[p(a_n) = r(a_n)\] Thus, the points $(a_n,p(a_n))$ are on the graph of the remainder. If all the roots of $d(x)$ are unique, then a system of equations can be made to find the remainder $r(x)$.

Examples

Introductory

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

Intermediate