Show that either $f$ is irreducible or $f$ factors into irreducible polynomials of degree $3$ over $K$. 
Let $f(x)$ be an irreducible polynomial of degree $6$ over field $K$.
If $L$ is a field extension of $K$ and $[L: K]=2$ then show that either $f$ is irreducible or $f$ factors into irreducible polynomials of degree $3$  over  $K$.

Attempt:
If $f$ is irreducible then we are done.
Otherwise $f$ factors into either $1+5$ or $2+4$ or $3+3$ degree polynomials.
I am unable to derive a contradiction for the first two cases.
Please give some hints.
 A: Proof that $f$ can't have a linear factor in $L$: Assume for contradiction that it has one, and let that linear factor be $x - \alpha$. Then $\alpha \in L$, and $f$ is the minimal polynomial of $\alpha$ over $K$. Can you get a contradiction from this?
The 2+4 works similarily, except we take the 2 factor, and find its roots (which might not be in $L$, but at the very least are contained in a degree 2 extension of $L$, which is then a degree 4 extension of $K$). Let $\beta$ be one of those roots, and derive a contradiction like above.
A: You might want to know the following general fact:

Let $F_1/F$ be field extension where $F_1$ is the splitting of some $f(x)\in F[x]$. For an irreducible $g(x)\in F[x]$, if
  $$g(x) = g_1 (x) g_2 (x) ... g_r (x) \quad \quad \text{ in } F_1[x]$$
  where each $g_i$ is irreducible in $F_1[x]$, then $\deg g_1 = \deg g_2 = \cdots = \deg g_r$.

In your example, $[L:K]=2$, so $L$ is a splitting field over $K$, the above result immediately addresses your concern.
