Returning back to your problem with in the field . Look at the first factor, , it has no zeros in by simply checking. Therefore this polynomial is irreducible. Form the factor ring which turns out being a field since is an irreducible polynomial, this is also a finite field with elements, so we will refer to it as . The mapping embeds in and therefore we can think of being contained in . Let and so , this means that is root of the polynomial . It is not hard to see that is another root of . Therefore, . Now the question is whether the second factor, , splits over . A little guessing shows that is a root of this polynomial because . And so the other root is easy to find which is which means and . Therefore, is a splitting field.