Determine whether $24x^5-30x^4+5=0$ is solvable by radicals over $ \mathbb{Q}?$ 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 $gal( \mathbb{Q}(\beta): \mathbb{Q})=?$
 A: It's a general fact that if $f(x)$ is an irreducible polynomial over $\mathbb{Q}$ of prime degree $p\geq 5$ having exactly $p-2$ real roots, then the Galois group of $f$ is $S_p$.
In this case $f(x)=24x^5-30x^4+5$ is irreducible by Eisenstein's criterion and Gauss's lemma. Also $f(-1)<0$, $f(0)>0$, $f(1)<0$, and $f(2)>0$, so $f$ has at least three real roots by the intermediate value theorem.
On the other hand, $f^{\prime}(x)=120x^3(x-1)$, which has exactly two roots. Hence $f(x)$ has at most three real roots by Rolle's theorem.
Thus $f(x)$ has exactly three real roots, hence the Galois group of $f$ over $\mathbb{Q}$ is $S_5$. And $S_5$ is not a solvable group, so $f$ isn't solvable by radicals.
A: 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 $x^5-x-16$). 
A: Using the IVT, we find there are three real roots in $\;(-1,0)\,,\,(0,1)\,,\,(1,2)\;$, and since the polynomial's derivative is $\;120x^3(x-1)\;$, the derivative is positive for $\;x<0\;$ or $\;x>1\;$ , which means that before the smallest root above, and after the greatest one there are no more real roots, and thus there is a conjugate pair $\;w,\,\overline w\;$ of complex non-real roots, which means the Galois group of the polynomial is $\;S_5\;$ (can see why?) and is thus non solvable by radicals.
