First you should have learned that roots in *are of the form If not tell me I'll show you.
Notice that it could be the case that none of them works. In this case the polynomial is irreducible over .
Else, let be a root found using the criterion .
Then it is possible to rewrite the polynomial as . You can now simply factor the second polynomial with the criterion .
At this point you can use the Eisenstein's criterion.
Even though you have divided these coefficients by some number, you can again put it in the form of a polynomial with integer coefficient by multiplying the whole thing by the least common multiple of the denominator.
You still have that some but that is some multiple of .
Therefore, this h(x) is irreducible and you thus are either one or no root.