Quartz 4

rational-root-theorem

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Rational Root Theorem

Summary: Any rational root of an integer-coefficient monic cubic must be an integer divisor of the constant term; Euler proves this from Vieta’s product-of-roots relation and from the impossibility of fractional denominators.

Sources: chapter-1.4.11, chapter-1.4.12, chapter-1.4.13

Last updated: 2026-05-03

Statement

For a monic cubic with integer coefficients , every rational root must be an integer divisor of (source: chapter-1.4.11, §722).

Proof via Vieta (§722)

Since the constant term equals the product of all three roots (see vieta-formulas), any root must divide . So when searching by trial, only the divisors of need be checked.

Proof via Denominators (§734)

Suppose (in lowest terms, ) is a root. Substituting into :

The first term has denominator ; all other terms have denominators , , or . Multiplied through by , the first term becomes an integer while the rest still contain factors of —a contradiction. Hence no fractional root can exist. If all integer divisors of fail, the roots are irrational (or imaginary) and Cardan’s formula must be used.

Practical Use

  1. Factor the constant term to list its divisors.
  2. Use the sign rule (cubic-equations) to decide whether to try positive or negative divisors.
  3. Substitute each candidate; if one satisfies the equation, divide by to reduce to a quadratic.
  4. If no divisor works, apply cardanos-rule.

For equations with fractional or non-unit leading coefficients, first clear denominators with or use when the constant term is 1 (source: chapter-1.4.11, §723–724).

Extension to quartics (§756)

The same argument holds for quartics: since the constant term (product of all four roots), any rational root of a monic integer-coefficient quartic must divide (source: chapter-1.4.13, §756). After finding a root , divide the quartic by to get a cubic solvable by cardanos-rule. Use descartes-sign-rule to determine which divisors (positive or negative) to test first.

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