Integer Polynomials and the Rational Roots Theorem

For all polynomials with integer coefficients, and , if is a rational root of with , then and .

(see GCD)

RRT is not if and only if, so it provides "candidates" for roots, but does not guarantee that those are roots

Find all rational roots of , and express as a product of irreducible polynomials in .

Using RRT, the divisors of are and , and the divisors of 3 are and . Hence, the candidates for rational roots are:

Now test each of these candidates

Thus, the only rational root is , so is a factor, and using long division, we obtain:

The quotient has no rational roots, since any rational root of must be a rational root of , and since , is not a root of . By factor theorem, has no linear factors.

Assume for the sake of contradiction, that is reducible in .
Then, from Degree of a Product, (which has degree 3) can be written as a product of two factors in , one of degree q and the other of degree 2.
Hence, has a linear factor, which is a contradiction.

We conclude that is irreducible in , so is an expression of using irreducible polynomials.