a factor of a n . When a n = , if there is a rational root p q , then as per theorem . q is a factor of a n , then we must have q = ± . Thus p must be an integer.
So a monic polynomial with integer coefficient cannot have non-integral rational roots. So when a n = , if at all there is a rational root, it must be an integer and the integer should divide a . (We say an integer a divides an integer b , if b ad for some integer d .) As an example let us consider the equation x . The divisors of are ± ± ± ± From Rational Root Theorem, we can conclude that ± ± ± ± are the only possible solutions of the equation.
It does not mean that all of them are solutions. The two values − and satisfy the equation and other values do not satisfy the equation. Moreover, if we consider the equation x , according to the Rational Root theorem, the possible solutions are ± ± ± ; but none of them is a solution. The Rational Root Theorem helps us only to guess a solution and it does not give a solution.
Example . Find the roots of According to our notations, a n = and a . If p q is a zero of the polynomial, then as ( , ) p q = must divide and q must divide . Clearly, the possible values of p are , , and the possible values of q are , , − .
Using these p and q we can form only the fractions - - ± , ± , ± , ± . Among these eight possibilities, after verifying by substitution, we get − is the only rational zero. To find other zeros, we divide the given polynomial