Rational root theorem

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 $r = \frac pq$ , $p|a_0$ 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 $P(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots +a_0$ , 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 $P(x)$ , $0=a_n\left(\frac{p}{q}\right)^n+\cdots +a_0.$ Multiplying by $q^n$ yields $0=a_np^n+a_{n-1}p^{n-1}q+\cdots+a_0q^n.$ Using modular arithmetic modulo $p$ , we have $a_0q^n\equiv 0\pmod p$ . As $q$ and $p$ are relatively prime, $p|a_0$ . Via similar logic in modulo $q$ , $q|a_n$ , which completes the proof. $\square$

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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