# Difference between revisions of "Rational Root Theorem"

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== Proof == | == Proof == | ||

− | Given <math>\frac{p}{q}</math> is a rational root of a polynomial <math>f(x)=a_nx^n+x_{n-1}x^{n-1}+\cdots +a_0</math>, we wish to show that <math>p|a_0</math> and <math>q|a_n</math>. Since <math>\frac{p}{q}</math> is a root, <cmath>0=a_n\left(\frac{p}{q}\right)^n+\cdots +a_0</cmath> Multiplying by <math>q^n</math>, we have: <cmath>0=a_np^n+a_{n-1}p^{n-1}q+\cdots+a_0q^n</cmath> Examining this in modulo <math>p</math>, we have <math>a_0q^n\equiv 0\pmod p</math>. As <math>q</math> and <math>p</math> are relatively prime, <math>p|a_0</math>. With the same logic, but with modulo <math>q</math>, we have <math>q|a_n</math>, and we are done. | + | Given <math>\frac{p}{q}</math> is a rational root of a polynomial <math>f(x)=a_nx^n+x_{n-1}x^{n-1}+\cdots +a_0</math>, where the coefficients are integers, we wish to show that <math>p|a_0</math> and <math>q|a_n</math>. Since <math>\frac{p}{q}</math> is a root, <cmath>0=a_n\left(\frac{p}{q}\right)^n+\cdots +a_0</cmath> Multiplying by <math>q^n</math>, we have: <cmath>0=a_np^n+a_{n-1}p^{n-1}q+\cdots+a_0q^n</cmath> Examining this in modulo <math>p</math>, we have <math>a_0q^n\equiv 0\pmod p</math>. As <math>q</math> and <math>p</math> are relatively prime, <math>p|a_0</math>. With the same logic, but with modulo <math>q</math>, we have <math>q|a_n</math>, and we are done. |

==Problems== | ==Problems== |

## Revision as of 23:51, 15 March 2012

*This article is a stub. Help us out by expanding it.*

Given a polynomial with integral coefficients, . The **Rational Root Theorem** states that if has a rational root with relatively prime positive integers, is a divisor of and is a divisor of .

As a consequence, every rational root of a monic polynomial with integral coefficients must be integral.

This gives us a relatively quick process to find all "nice" roots of a given polynomial, since given the coefficients we have only a finite number of rational numbers to check.

## Contents

## Proof

Given is a rational root of a polynomial , where the coefficients are integers, we wish to show that and . Since is a root, Multiplying by , we have: Examining this in modulo , we have . As and are relatively prime, . With the same logic, but with modulo , we have , and we are done.

## Problems

### Easy

Factor the polynomial .

### Intermediate

Find all rational roots of the polynomial .

Prove that is irrational, using the Rational Root Theorem.