Hello,
I know nothing of Galois theory, so I'm not too sure you're looking for rational solutions, real solutions, or complex solutions... (my guess goes for rational zeroes ^^)
Anyway, this may help : The Rational Roots Test
Hi moo,
looking for rational solutions would have given the answer for a polynomial in of degree 2 or 3 (in such case, "no solution irreducibility"). With higher degrees, the polynomial can be a product of irreducible polynomials and be reducible without having ant root. But thanks
If you use the rational zeros theorem you will realize it has no zeros. Therefore, if this polynomial is reducible then it means where . Expand RHS and compare coefficients and show that is impossible. This is a really painful way of proving irreducibility! Here is an easier way. Replace by and then . Now apply Einsteinstein irreducibility test with .
I do not want to be insulting, but how many times do I have to tell you that not having zeros does not imply irreducibility! I think that is my third time telling you this already.
Stupid me >< I always miss that !