TITLE: Determine whether $24x^5-30x^4+5=0$ is solvable by radicals over $ \mathbb{Q}?$
QUESTION [5 upvotes]: Determine whether $24x^5-30x^4+5=0$ is solvable by radicals over $ \mathbb{Q}$
My try:There is a theorem that a polynomial can be solved by radicals if and only if its Galois group is a solvable group.here the polynomial is irreducible in $ \mathbb{Q}$ by Eisenstein's criteria.if $\beta$ is the root of the polynomial then $gal( \mathbb{Q}(\beta): \mathbb{Q})=?$

REPLY [2 votes]: I learnt this technique recently on MSE. It is based on the general theorem of Galois:
Theorem: If $p(x) $ is an irreducible polynomial of prime degree with rational coefficients then it is solvable by radicals if and only if all the roots of this polynomial can be expressed as rational functions of any two of its roots.
This has a nice corollary that if an irreducible polynomial of prime degree is solvable by radicals and has two real roots then all the other roots being rational functions of these roots must be real. So we have
Corollary: If $p(x) $ is an irreducible polynomial of prime degree with rational coefficients and it is solvable by radicals then either it has one real root or it has all its roots real.
This handles many polynomials of prime degree and your example also fits here because it has only three real roots and thus is not solvable by radicals. The tough case is when such polynomial has only one real root and yet it may happen that it is not solvable by radicals (for example $x^5-x-16$).