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 a_{0} 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
Thus, a list of possible rational roots of the equation can be derived using the formula .
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.
Contents
Proof
Let P(x) = a_{n}x^{n} + a_{n1}x^{n1} + ... + a_{1}x + a_{0} for some a_{0}, ..., a_{n} ∈ Z, and suppose P(p/q) = 0 for some coprime p, q ∈ Z:
Shifting the constant term and multiplying by q^{n},
Shifting the leading term and multiplying by q^{n},
All terms in these equations are integers, which implies p  a_{0}q^{n} and q  a_{n}p^{n}. But p, q^{n} and q, p^{n} are coprime. Hence, by Euclid's lemma, p  a_{0} and q  a_{n}.^{[1]}
Example
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 using the Horner scheme (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. 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 t^{3} remains the same as the coefficient of x^{3}. 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 r_{1} is found, the Horner scheme will also yield a polynomial of degree n − 1 whose roots, together with r_{1}, 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 a_{0}, then 0 is one of the rational roots of the equation.
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