When is a 5th degree polynomial with at least 1 non-real root solvable by radicals? Let $f(X)$ be an irreducible polynomial of degree 5 with coefficents in the field of rational numbers $\mathbb{Q}$. Assume that $f$ has at least one non-real root in the complex field $\mathbb{C}$. Assume further that the discriminant of $f$ is a square in $\mathbb{Q}$.

Let $r$ be a root of $f$, and let $K$ be the field $\mathbb{Q}(r)$, so that $f$ factors in $K[X]$ as $$f = (X − r)g$$ with $g$ of degree 4. Prove that $f$ is solvable by radicals if and only if $g$ is reducible in $K[X]$.

My thoughts:
I know that since the discriminant of $f$ is a square in $\mathbb{Q}$, $G\subset A_5,$ and since $f$ is irreducible, $G$ must be transitive. So we must have that either $G\cong A_5$ or $G\cong D_5$ (the dihedral group of order 10). Moreover, since char $\mathbb{Q}=0,$ we know that $f$ is solvable by radicals if and only if $G$ is solvable.
Since $A_5$ is simple and non-abelian, $A_5$ is not solvable. On the other hand, $D_5$ is solvable: $D_5$ has an element $r$ with $r^5=e$, and $[D_5 : \left<r\right>]=2,$ so $\left<r\right>\triangleleft D_5.$ Now 
$$D_5\triangleright\left<r\right>\triangleright\{e\}.$$  So it suffices to show that $g$ is reducible in $K[X]$ if and only if $G\cong D_5.$ 
If $f$ has exactly 2 non-real roots, then the only nontrivial $\mathbb{Q}$-automorphism is complex conjugation, so $G$ contains a transposition, which implies $G\cong S_5$. Thus, $f$ must have exactly one real root and 4 non-real roots.
This is where I'm stuck. How can we relate the reducibility of $g$ in $K[X]$ to the Galois group of $f$?
 A: HINT: The subgroup of $G$ fixing $r$ acts transitively on the remaining roots iff $g$ is irreducible.
A: Another way to think the problem is the following.
First, show that $A_5$ cannot have a subgroup of size $20.(1)$
If such subgroup $H$ existed, $H$ would be a subgroup of $A_5$ of index $3$, which  cannot be normal, why would this imply $A_5$ to have a normal subgroup of index $2$?; consider the action of $G$ on the cosets of $H$.
Let $F$ be the splitting field of the polynomial $f$. Then $[F:\Bbb Q]=|Gal(F/\Bbb Q)|$.$(2)$
If $g$ is irreducible in $K[X]$, we get $20\mid[F:\Bbb Q]$. By $(2)$ we have  $[F:\Bbb Q]\mid |A_5|$, because of $(1)$ we must have that $Gal(F/\Bbb Q)=A_5$(why?), hence $f$ is not soluble by radicals.
If $g$ is reducible in $K[X]$, there are two cases:


*

*Suppose $g=h_1h_2$, where $h_1$ and $h_2$ are irreducible over $K[X]$ both of degree $2$. Then we must have that either $[F:\Bbb Q]=20$ or $[F:\Bbb Q]=10$(why?), by what we saw above the first case cannot happen, so that $|Gal(F/\Bbb Q)|=10$, and any group of this size must be soluble; use Sylow's third theorem.

*If $g=h_1h_2$, with $h_1$ irreducible of degree $3$, let $\gamma$ be a root of $h_1$. We claim that $F=K[\gamma]$. Otherwise, we'd get $[F:\Bbb Q]=30$(why?), so that $|Gal(F/\Bbb Q)|=30$, however as $A_5$ is simple of size $60$, it cannot have subgroups of size $30$. Thus $[F:\Bbb Q]=15$, i.e., $|Gal(F/\Bbb Q)|=15$, hence $Gal(F/\Bbb Q)$ is soluble.

