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 $gal( \mathbb{Q}(\beta): \mathbb{Q})=?$
Best Answer
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.