# Difference between revisions of "Rational Root Theorem"

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1. <math>(x-4)(x-2)(x+1)</math> | 1. <math>(x-4)(x-2)(x+1)</math> | ||

− | 2. <math>\text{ | + | 2. <math>\text{There are no rational roots for the polynomial.} </math> |

3. A polynomial with integer coefficients and has a root as <math>\sqrt{2}</math> must also have <math>-\sqrt{2}</math> as a root. The simplest polynomial is <math>(x+\sqrt{2})(x-\sqrt{2})</math> which is <math>x^2-2=0</math>. We see that the only possible rational roots are <math>\pm 1</math> and <math>\pm 2</math>, and when substituted, none of these roots work. | 3. A polynomial with integer coefficients and has a root as <math>\sqrt{2}</math> must also have <math>-\sqrt{2}</math> as a root. The simplest polynomial is <math>(x+\sqrt{2})(x-\sqrt{2})</math> which is <math>x^2-2=0</math>. We see that the only possible rational roots are <math>\pm 1</math> and <math>\pm 2</math>, and when substituted, none of these roots work. | ||

{{stub}} | {{stub}} |

## Latest revision as of 15:18, 13 February 2020

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 's 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 , which completes the proof.

## Problems

### Easy

1. Factor the polynomial .

### Intermediate

2. Find all rational roots of the polynomial .

3. Prove that is irrational, using the Rational Root Theorem.

### Answers

1.

2.

3. A polynomial with integer coefficients and has a root as must also have as a root. The simplest polynomial is which is . We see that the only possible rational roots are and , and when substituted, none of these roots work.
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