Let $L$ be an extension of a field $F$ with $[L:F]=2$. Let $f(x)\in F[x]$ be irreducible.
Show that either $f(x)$ is irreducible in $L[x]$ or $f(x)=g(x)h(x)$ where $g(x), h(x)$ are irreducible in $L[x]$ and $\deg(g)=\deg(h)$.
What I tried is to suppose $f(x)$ is not irreducible in $L[x]$ first. Then $f(x)=g(x)h(x)$ for some $g(x), h(x)\in L[x]$.
Then perhaps try to suppose to the contrary $g(x)=p(x)q(x)$, for some nonconstant $p(x),q(x)\in L[x]$.
I know that the roots of $f$ are certainly not in $F$, but I can't say for certainty are they in $L$?
Thanks for any help. I am quite stuck here.
The galois group has two elements and its action on the irreducible factors of $f$ is transitive. Hence there are at most two irreducible factors and if there are two of them, they must have the same degree, since the are swapped by the non-trivial element of the galois group.
If the extension is not Galois, we are in char $2$ and $L=F(a)$ with $a^2 \in F$. Then any square of a polynomial in $L[x]$ lies in $F[x]$. From here you can deduce that $f$ remains irreducible over $L$ or is a square of an irreducible polynomial.