Rational root theorem

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

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.

Thus, a list of possible rational roots of the equation can be derived using the formula .

For example, every rational solution of the equation

must be among the numbers symbolically indicated by

±

which gives the list of possible answers:

These root candidates can be tested, for example using the Horner scheme. 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. 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:

Therefore,

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, the Horner scheme 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. The fundamental theorem of algebra states that any polynomial with integer (or real, or even complex) coefficients must have at least one root in the set of complex numbers. Since the roots of a polynomial with real coefficients occur in complex conjugate pairs, and a real number is self-conjugate, any polynomial of odd degree (the degree being n = 3 in the example above) with real coefficients must have a root in the real numbers.

If the equation lacks a constant term a₀, then 0 is one of the rational roots of the equation.

The 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.

A Proof

Suppose p/q is a root to our integer coefficient polynomial, f(x), of degree n, where p and q are integers. Suppose p/q is fully reduced so that p and q are co-prime (i.e., the greatest common divisor of p and q is 1; in short gcd(p, q) = 1).

Then,

ƒ(p/q) = 0.

Multiplying both sides of this by qⁿ⁻¹, yields

qⁿ⁻¹ · ƒ(p/q) = a_n · pⁿ / q + {sum of integers} = 0.

Hence, a_n · pⁿ / q is an integer. Because p and q are co-prime, q divides a_n.

Similarly, suppose we had multiplied by qⁿ/p instead. Then we have

qⁿ · ƒ(p/q) / p = {sum of integers} + a_o · qⁿ / p = 0.

Rearranging, we find a_o · qⁿ / p is an integer. Because p and q are co-prime p divides a_o. Thus p is an integer factor of the constant term, q is an integer factor of the leading coefficient, and p/q is a root of ƒ(x).

External links

• Another proof that n^th roots of integers are irrational, except for perfect nth powers by Scott E. Brodie