---
tag: 0819CQ1V8H
kind: statement
role: theorem
name: Rational root theorem
aliases: [rational-root-test]
msc: ["12E05"]
uses: [00EHX6FWC0]
sources:
  - url: https://en.wikipedia.org/wiki/Rational_root_theorem
    locator: lead section
same_as:
  wikidata: Q180345
  wikipedia: 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|00EHX6FWC0]] 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|06YMXJMXYS]] 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.