abstract algebra - Splitting field and dimension of irreducible polynomials

Given a field extension $L/K$, $\alpha, \beta \in L$ and $f,g \in K[x]$ irreducible polynomials with $f(\alpha)=g(\beta)=0$. Then

$$\operatorname{dim}_K(K(\alpha,\beta)) = \deg(f) \cdot \operatorname{dim}_{K(\alpha)}(K(\alpha,\beta)) = \deg(g) \cdot \operatorname{dim}_{K(\beta)}(K(\alpha,\beta))$$

holds. I have no idea how to prove that as the degree of the polynomial because the dimension of the splitting field has to only a divisor of the degree but not the exact product.

$f \in K(\beta)[x]$ is ireducible iff $g \in K(\beta)[x]$ is irreducible. Any ideas how to prove that?

Answer

Hints: the first hint is what Gregor wrote in the comment (Hagen gave a hint on the proof), the second hint that if $K/F$ is a field extension and $K=F(\alpha)$ where $\alpha$ is algebric over $F$ then $[K:F]=deg(m_{\alpha,F})$ also keep in mind my answer from before that this is the degree of any other irreducible polynomial $p$ that $p(\alpha)=0$