Why is $f(X)=X^5-X^3-5X^2+5$ is solvable by radicals?

Question

I am trying to understand solvability by radicals through examples.

Can anyone demonstrate to me why:

• $f \in \mathbb Q[X]$ where $f(X)=X^5-X^3-5X^2+5$ is soluble by radicals

but

• $g \in \mathbb Q[X]$ where $g(X)=X^5-5X^4+10X^3+50X^2-5$ is not solvable by radicals

I understand $f$ can be reducible to $(X-1)(X+1)(X^3-5)$ and these four distinct roots imply solubility?

Sorry I'm still trying to get my head around this concept and I thought examining a solvable and non-solvable example would assist my understanding.

Any help would be amazing!

Answer 1

There's a theorem within Galois theory (you will find it, for instance, in Ian Stewart's Galois Theory) that says that, given a prime number $p$, if a polynomial $P(x)\in\mathbb{Q}[x]$ with degree $p$ has $2$ and only $2$ non-real roots, then its Galois group is $S_p$. That's the case here, with $p=5$ and $P(x)=g(x)$ (whose non-real roots are about $3.51966\pm3.46792i$). Since the Galois group of $g(x)$ is $S_5$, which is not a solvable group, the equation $g(x)=0$ is not solvable by radicals.

Note: I am assuming that $g(x)$ is irreducible in $\mathbb{Q}[x]$; I did not check it.