However does do the job because it divides all of the coefficients and
Therefore it is irreducible.
once again for b) You have used a prime that is not allowed and works.
For the last one Since both 2 and 5 are relatively prime the theorem does not apply directly. As suggested by the hint:
Now if we let the polynomial is irreducible. So this implies that is irreducible.
Lets see why. Suppose that is irreduicble but is then for , but then
but now is reducible and this is a contradiction.