Rational root theorem

Statement

Let $f = a_n x^n + a_{n-1} x^{n-1} + \dots + a_0 \in \mathbb{Z}[x]$ with $n \ge 1$ and $a_n \neq 0$. Let $r = p/q$ be a rational root of $f$, with $p, q \in \mathbb{Z}$, $q > 0$ and $\gcd(p, q) = 1$. Then $p$ divides $a_0$ and $q$ divides $a_n$.

Remarks

The proof: multiply $f(p/q) = 0$ by $q^n$. Then $a_n p^n = -q(\dots)$ and $a_0 q^n = -p(\dots)$, and $\gcd(p, q) = 1$ gives the two divisibilities. For example, the only possible rational roots of $2x^3 - 3x + 1$ are $\pm 1$ and $\pm \tfrac{1}{2}$. Evaluation shows that $1$ is a root. The same statement holds over every unique factorisation domain and its field of fractions. For a monic $f$, every rational root is an integer.

Explanations

L1 human

The theorem gives a finite list of candidates for the rational roots. You test each candidate by evaluation. For $x^3 - 2$, the candidates are $\pm 1$ and $\pm 2$, and none is a root, so $\sqrt[3]{2}$ is irrational.