Difference between revisions of "Rational root theorem"

Revision as of 13:05, 15 July 2021

In algebra, the rational root theorem states that given an integer polynomial $P(x)$ with leading coefficent $a_n$ and constant term $a_0$ , if $P(x)$ has a rational root in lowest terms $r = \frac pq$ , then $p|a_0$ and $q|a_n$ .

This theorem aids significantly at finding the "nice" roots of a given polynomial, since the coefficients entail only a finite amount of rational numbers to check as roots.

Proof

Let $\frac{p}{q}$ be a rational root of $P(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_0$ , where all $a_r$ are integers; we wish to show that $p|a_0$ and $q|a_n$ . Since $\frac{p}{q}$ is a root of $P(x)$ , $0 = a_n \left(\frac{p}{q}\right)^n + a_{n-1} \left(\frac{p}{q}\right)^{n-1} + \cdots + a_1 \left(\frac{p}{q}\right) + a_0.$ Multiplying by $q^n$ yields $0 = a_n p^n + a_{n-1} p^{n-1} q + \cdots + a_1 p * q^{n-1} + a_0 q^n.$ Using modular arithmetic modulo $p$ , we have $a_0 q^n \equiv 0\pmod p$ , which implies that $p | a_0 q^n$ . Because we've defined $p$ and $q$ to be relatively prime, $\gcd(q^n, p) = 1$ , which implies $p | a_0$ by Euclid's lemma. Via similar logic in modulo $q$ , $q|a_n$ , as required. $\square$

Problems

Here are some problems that are cracked by the rational root theorem. The answers can be found here.

Introductory

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

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.

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.

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-In [[algebra]], the '''rational root theorem''' states that given an integer [[polynomial]] <math>P(x)</math> with leading coefficent <math>a_n</math> and constant term <math>a_0</math>, if <math>P(x)</math> has a rational root in lowest terms <math>r = \frac pq</math>, <math>p|a_0</math> and <math>q|a_n</math>.
+In [[algebra]], the '''rational root theorem''' states that given an integer [[polynomial]] <math>P(x)</math> with leading coefficent <math>a_n</math> and constant term <math>a_0</math>, if <math>P(x)</math> has a rational root in lowest terms <math>r = \frac pq</math>, then <math>p|a_0</math> and <math>q|a_n</math>.
 This theorem aids significantly at finding the "nice" roots of a given polynomial, since the coefficients entail only a finite amount of rational numbers to check as roots.
 == Proof ==
-Let <math>\frac{p}{q}</math> be a rational root of <math>P(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots +a_0</math>, where all <math>a_r</math> 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 of <math>P(x)</math>, <cmath>0=a_n\left(\frac{p}{q}\right)^n+\cdots +a_0.</cmath> Multiplying by <math>q^n</math> yields <cmath>0=a_np^n+a_{n-1}p^{n-1}q+\cdots+a_0q^n.</cmath> Using [[modular arithmetic]] 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>. Via similar logic in modulo <math>q</math>, <math>q|a_n</math>, which completes the proof. <math>\square</math>
+Let <math>\frac{p}{q}</math> be a rational root of <math>P(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_0</math>, where all <math>a_r</math> 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 of <math>P(x)</math>, <cmath>0 = a_n \left(\frac{p}{q}\right)^n + a_{n-1} \left(\frac{p}{q}\right)^{n-1} + \cdots + a_1 \left(\frac{p}{q}\right) + a_0.</cmath> Multiplying by <math>q^n</math> yields <cmath>0 = a_n p^n + a_{n-1} p^{n-1} q + \cdots + a_1 p * q^{n-1} + a_0 q^n.</cmath> Using [[modular arithmetic]] modulo <math>p</math>, we have <math>a_0 q^n \equiv 0\pmod p</math>, which implies that <math>p | a_0 q^n</math>. Because we've defined <math>p</math> and <math>q</math> to be relatively prime, <math>\gcd(q^n, p) = 1</math>, which implies <math>p | a_0</math> by [[Euclid's lemma]]. Via similar logic in modulo <math>q</math>, <math>q|a_n</math>, as required. <math>\square</math>
 == Problems ==

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