Difference between revisions of "Rational Root Theorem"

m (Answers)
m (Proof)
(5 intermediate revisions by 4 users not shown)
Line 1: Line 1:
{{stub}}
+
 
  
  
Line 10: Line 10:
 
== 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>, where the <math>a_n</math>'s 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>, which completes the proof.
+
Given <math>\frac{p}{q}</math> is a rational root of a polynomial <math>f(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots +a_0</math>, where the <math>a_n</math>'s 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>, which completes the proof.
  
 
==Problems==
 
==Problems==
Line 30: Line 30:
  
 
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}}

Revision as of 21:31, 22 March 2021


Given a polynomial $P(x) = a_n x^n + a_{n - 1}x^{n - 1} + \ldots + a_1 x + a_0$ with integral coefficients, $a_n \neq 0$. The Rational Root Theorem states that if $P(x)$ has a rational root $r = \pm\frac pq$ with $p, q$ relatively prime positive integers, $p$ is a divisor of $a_0$ and $q$ is a divisor of $a_n$.

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.

Proof

Given $\frac{p}{q}$ is a rational root of a polynomial $f(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots +a_0$, where the $a_n$'s are integers, we wish to show that $p|a_0$ and $q|a_n$. Since $\frac{p}{q}$ is a root, \[0=a_n\left(\frac{p}{q}\right)^n+\cdots +a_0\] Multiplying by $q^n$, we have: \[0=a_np^n+a_{n-1}p^{n-1}q+\cdots+a_0q^n\] Examining this in modulo $p$, we have $a_0q^n\equiv 0\pmod p$. As $q$ and $p$ are relatively prime, $p|a_0$. With the same logic, but with modulo $q$, we have $q|a_n$, which completes the proof.

Problems

Easy

1. Factor the polynomial $x^3-5x^2+2x+8$.

Intermediate

2. Find all rational roots of the polynomial $x^4-x^3-x^2+x+57$.

3. Prove that $\sqrt{2}$ is irrational, using the Rational Root Theorem.

Answers

1. $(x-4)(x-2)(x+1)$

2. $\text{There are no rational roots for the polynomial.}$

3. A polynomial with integer coefficients and has a root as $\sqrt{2}$ must also have $-\sqrt{2}$ as a root. The simplest polynomial is $(x+\sqrt{2})(x-\sqrt{2})$ which is $x^2-2=0$. We see that the only possible rational roots are $\pm 1$ and $\pm 2$, and when substituted, none of these roots work. This article is a stub. Help us out by expanding it.