The theorem is used to find all rational roots of a polynomial, if any. It gives a finite number of possible fractions which can be checked to see if they are roots. If a rational root is found, a linear polynomial can be factored out of the polynomial using polynomial long division, resulting in a polynomial of lower degree whose roots are also roots of the original polynomial.
Cubic equation
The general cubic equation with integer coefficients has three solutions in the complex plane. If the rational root test finds no rational solutions, then the only way to express the solutions algebraically uses cube roots. But if the test finds a rational solution, then factoring out leaves a quadratic polynomial whose two roots, found with the quadratic formula, are the remaining two roots of the cubic, avoiding cube roots.
Proofs
First proof
Let Suppose for some coprime : Now multiply both sides by. By shifting the constant term to the right side, and factoring out on the left side, produces Thus, divides. But is coprime to and therefore to, so by Euclid's lemma must divide the remaining factor of the product. On the other hand, shifting the leading term to the right side and factoring out on the left side, gives Reasoning as before, it follows that divides.
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, then it also factors in as a product of primitive polynomials. Now any rational root corresponds to a factor of degree 1 in of the polynomial, and its primitive representative is then, assuming that and are coprime. But any multiple in of has leading term divisible by and constant term divisible by, which proves the statement. This argument shows that more generally, any irreducible factor of can be supposed to have integer coefficients, and leading and constant coefficients dividing the corresponding coefficients of .
Examples
First
In the polynomial any rational root fully reduced would have to have a numerator that divides evenly into 1 and a denominator that divides evenly into 2. Hence the only possible rational roots are ±1/2 and ±1; since neither of these equates the polynomial to zero, it has no rational roots.
Second
In the polynomial the only possible rational roots would have a numerator that divides 6 and a denominator that divides 1, limiting the possibilities to ±1, ±2, ±3, and ±6. Of these, 1, 2, and –3 equate the polynomial to zero, and hence are its rational roots.
Third
Every rational root of the polynomial must be among the numbers symbolically indicated by: These 8 root candidates can be tested by evaluating, for example using Horner's method. It turns out there is exactly one with. This process may be made more efficient: if, it can be used to shorten the list of remaining candidates. For example, does not work, as. Substituting yields a polynomial in with constant term, while the coefficient of remains the same as the coefficient of. Applying the rational root theorem thus yields the possible roots, so that True roots must occur on both lists, so list of rational root candidates has shrunk to just and If rational roots are found, Horner's method will also yield a polynomial of degree whose roots, together with the rational roots, are exactly the roots of the original polynomial. If none of the candidates is a solution, there can be no rational solution.
