Rational root theorem

See also: Eisenstein criterion

In algebra, the rational root theorem (or rational root test, rational zero theorem, rational zero test or p/q theorem) states a constraint on rational solutions (or roots) of a polynomial equation

$a_nx^n+a_{n-1}x^{n-1}+\cdots+a_0 = 0\,\!$

with integer coefficients.

If a₀ and a_n are nonzero, then each rational solution x, when written as a fraction x = p/q in lowest terms (i.e., the greatest common divisor of p and q is 1), satisfies

• p is an integer factor of the constant term a₀, and
• q is an integer factor of the leading coefficient a_n.

The rational root theorem is a special case (for a single linear factor) of Gauss's lemma on the factorization of polynomials. The integral root theorem is a special case of the rational root theorem if the leading coefficient a_n = 1.

Proofs

A proof

Let P(x) = a_nxⁿ + a_n−1xⁿ⁻¹ + ... + a₁x + a₀ for some a₀, ..., a_n ∈ Z, and suppose P(p/q) = 0 for some coprime p, q ∈ Z:

$P\left(\tfrac{p}{q}\right) = a_n\left(\tfrac{p}{q}\right)^n + a_{n-1}\left(\tfrac{p}{q}\right)^{n-1} + \cdots + a_1\left(\tfrac{p}{q}\right) + a_0 = 0.$

If we multiply both sides by qⁿ, shift the constant term to the right hand side, and factor out p on the left hand side, we get

$\qquad p(a_np^{n-1} + a_{n-1}qp^{n-2} + \cdots + a_1q^{n-1}) = -a_0q^n.$

We see that p times the integer quantity in parentheses equals −a₀qⁿ, so p divides a₀qⁿ. But p is coprime to q and therefore to qⁿ, so by (the generalized form of) Euclid's lemma it must divide the remaining factor a₀ of the product.

If we instead shift the leading term to the right hand side and factor out q on the left hand side, we get

$\qquad q(a_{n-1}p^{n-1} + a_{n-2}qp^{n-2} + \cdots + a_0q^{n-1}) = -a_np^n.$

And for similar reasons, we can conclude that q divides a_n.^[1]

Proof using Gauss's lemma

Should there be a nontrivial factor dividing all the coefficients of the polynomial, then one can divide by the greatest common divisor of the coefficients so as to obtain a primitive polynomial in the sense of Gauss's lemma; this does not alter the set of rational roots and only strengthens the divisibility conditions. That lemma says that if the polynomial factors in ℚ[X], then it also factors in ℤ[X] as a product of primitive polynomials. Now any rational root p/q corresponds to a factor of degree 1 in ℚ[X] of the polynomial, and its primitive representative is then qx − p, assuming that p and q are coprime. But any multiple in ℤ[X] of qx − p has leading term divisible by q and constant term divisible by p, which proves the statement. This argument shows that more generally, any irreducible factor of P can be supposed to have integer coefficients, and leading and constant coefficients dividing the corresponding coefficients of P.

Example

For example, every rational solution of the equation

$3x^3 - 5x^2 + 5x - 2 = 0\,\!$

must be among the numbers symbolically indicated by

$\pm\frac{1,2}{1,3}\,,$

which gives the list of 8 possible answers:

$\pm\{1, 2, \frac{1}{3}, \frac{2}{3}\} .$

These root candidates can be tested using the Horner's method (for instance). In this particular case there is exactly one rational root. If a root candidate does not satisfy the equation, it can be used to shorten the list of remaining candidates.^[2] For example, x = 1 does not satisfy the equation as the left hand side equals 1. This means that substituting x = 1 + t yields a polynomial in t with constant term 1, while the coefficient of t³ remains the same as the coefficient of x³. Applying the rational root theorem thus yields the following possible roots for t:

$t=\pm\frac{1}{1,3}$

Therefore,

$x = 1+t = 2, 0, \frac{4}{3}, \frac{2}{3}$

Root candidates that do not occur on both lists are ruled out. The list of rational root candidates has thus shrunk to just x = 2 and x = 2/3.

If a root r₁ is found, Horner's method will also yield a polynomial of degree n − 1 whose roots, together with r₁, are exactly the roots of the original polynomial. It may also be the case that none of the candidates is a solution; in this case the equation has no rational solution. If the equation lacks a constant term a₀, then 0 is one of the rational roots of the equation.

See also

• Descartes' rule of signs
• Gauss–Lucas theorem
• Properties of polynomial roots
• Content (algebra)

Notes

1. D. Arnold, G. Arnold (1993). Four unit mathematics. Edward Arnold. pp. 120–121. ISBN 0-340-54335-3.
2. King, Jeremy D. "Integer roots of polynomials", Mathematical Gazette 90, November 2006, 455-456.

References

• Charles D. Miller, Margaret L. Lial, David I. Schneider: Fundamentals of College Algebra. Scott & Foresman/Little & Brown Higher Education, 3rd edition 1990, ISBN 0-673-38638-4, pp. 216–221
• Phillip S. Jones, Jack D. Bedient: The historical roots of elementary mathematics. Dover Courier Publications 1998, ISBN 0-486-25563-8, pp. 116–117 (online copy, p. 116, at Google Books)
• Ron Larson: Calculus: An Applied Approach. Cengage Learning 2007, ISBN 978-0-618-95825-2, pp. 23–24 (online copy, p. 23, at Google Books)

External links

• Weisstein, Eric W., "Rational Zero Theorem", MathWorld.
• RationalRootTheorem at PlanetMath
• Another proof that n^th roots of integers are irrational, except for perfect nth powers by Scott E. Brodie
• The Rational Roots Test at purplemath.com