If $G$ is a finite group with $\text{Aut}(G)$ solvable, then $G$ itself must be solvable. Because then $G/Z(G)$ is solvable and obviously $Z(G)$ is solvable, and hence $G$ itself must be solvable. But what about the converse? Can $G$ be solvable and $\text{Aut}(G)$ be non-solvable?

This is not for homework, but I would just like a hint please. The question asks If $f(x) \in \mathbb{F}[x]$ has degree $n$, show that there exists an extension field $\mathbb{E}$ of $\mathbb{F}$ such that $\mathbb{E}$ contains all the roots of $f(x)$. I believe induction is the way to go ...