Reducibility of $f \in \mathbb{Q}(\alpha)[x]$ for $[\mathbb{Q}(\alpha):\mathbb{Q}] \geq 2$ and $f \in \mathbb{Q}[x]$ Irreducible?

Question

I'm struggling to make any progress on the following problem.

Let $K = \mathbb{Q}(\alpha)$ be an algebraic extension of $\mathbb{Q}$ with $[\mathbb{Q}(\alpha):\mathbb{Q}]=m$, and suppose that $f \in \mathbb{Q}[x]$ is irreducible.

a. Suppose that $m=2$. Prove that either $f \in K[x]$ is still irreducible, or that $f=gh$ for $1 < \deg(g),\deg(h)$ and $\deg(g)=\deg(h)$.

b. Suppose now that $m > 2$. Does it follow that $f \in K[x] $ remains irreducible or factors into $m$ irreducible polynomials?

So I believe the answer to both is yes, but I'm not really sure how. For part a, I start by supposing that $f(x) = p_1(x)p_2(x)\dots p_r(x)$ for $p_i \in K[x]$. I want to show that the only possibility is that $r=2$, and that $\deg(p_1)=\deg(p_2)$. I'm aware of the identification $K \cong \mathbb{Q}[x] / \langle \text{Irr}_{\mathbb{Q}}(\alpha,x) \rangle$, and so one can think of the coefficients on each $p_i$ as being degree $1$ polynomials in $\mathbb{Q}[x] / \langle \text{Irr}_{\mathbb{Q}}(\alpha,x)\rangle$, but I have no idea what else to try.

For part b, I'm assuming that If one figures out part a properly, then part b follows by some basic inductive argument.

Any help would be great. Thanks!