Difference between revisions of "Rational Root Theorem"

Revision as of 01:32, 30 December 2008

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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 $f(x)=a_nx^n+x_{n-1}x^{n-1}+\cdots +a_0$ , we wish to show that $p|a_0$ and $q|a_n$ . Since $\frac{p}{q}$ is a root, $0=a_n(\frac{p}{q})^n+\cdots +a_0$ 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 $q|a_n$ , and we are done.

Problems

Intermediate

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

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

@@ Line 7: / Line 7: @@
 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 <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(\frac{p}{q})^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==

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Difference between revisions of "Rational Root Theorem"

Revision as of 01:32, 30 December 2008

Proof

Problems

Intermediate