Question 5 Hungerford algebra Pg 252 Consider the following problem :
If char K $\neq $2 and $f\in  K[x]$ is a cubic whose discriminant is a square in K, then f
is either irreducible or factors completely in K.
This question is from section "Galois Group of the Polynomial "  although I have read the section carefully but I am still not able to figure out which result should I use to solve the problem. So ,there is a result that says that Galois Group must be $S_3$ or $A_3$ and if it is $A_3$ then discriminant is equal to square of an element of K.
But Still I am not close to proving why it is either irreducible or factors completely.
So, Can you please help me with that?
 A: By using the assumption $\operatorname{char} K\ne2$ show the claim is true in the case where $f$ is inseparable. So we may assume that $f$ is separable. Let $F$ be a splitting field of $f$. Since the discriminant of $f$ is a square in $K$ the Galois group $G=G(F/K)$ is contained in $A_3$ (Here we again used $\operatorname{char}K\ne2$). In particular we have $\vert G\vert = 1$ or $\vert G\vert =3$. Now if $f$ is not irreducible argue why the case $\vert G\vert =3$ cannot happen. We then get $\vert G\vert=1$, i.e $F=K$.
Edit: Assume that $f$ is inseparable and reducible. So over some extension field $f$ factors as $(x-\alpha)(x-\beta)^2$ where at least one of $\alpha,\beta$ is already in $K$. If $\alpha=\beta$ we are done (since all linear factors are in $K$), so assume $\alpha\ne\beta$. If $\beta\in K$, then $(x-\beta)^2\in K[x]$, so we can divide $f$ byy $(x-\beta)^2$ over $K$ and get $(x-\alpha)\in K[x]$, so $\alpha\in K$. Assume $\alpha\in K$. Again by division we get $(x-\beta)^2\in K[x]$. In characteristic $p$ an irreducible inseparable polynomial always has degree $p^r$. Since we are not in characteristic $2$ this means that $(x-\beta)^2$ cannot be irreducible, hence $\beta\in K$.
In total we proved that if $f$ is inseparable and reducible then $f$ splis into linear factors over $K$.
