Let
be a polynomial in which has no rational root.
Show that is irreducible in
if the cubic resolvent
has no rational root.
If the cubic resolvent has no rational roots then it is irreducible. Therefore, the Galois group of must contain and so it splitting field is at least a -th degree extension. Now if the original polynomial was reducible then its will factor as a product of two quadradics which would mean its Galois group would be at most a 4 degree extension. Which is a contradiction.