If $p(x)$ is the minimal polynomial of $\alpha$, and $f(x) \in F[x]$, $f(\alpha) = 0$, Then $p(x) | f(x)$

Question

If $p(x)$ is the minimal polynomial of $\alpha$, and $f(x) \in F[x]$, $f(\alpha) = 0$,

Show:

$$p(x) | f(x)$$

I'm not sure if it's a direct consequence of the proof that minimal polynomial of an algebraic element over a field is irreducible. Any help is appreciated.

The proof:

A minimal polynomial is irreducible. Let $E/F$ be a field extension over $F$ as above, $\alpha \in E$, and $f \in F[x]$ a minimal polynomial for $\alpha$.

Suppose $f = gh$, where $g, h ∈ F[x]$ are of lower degree than $f$. Now $f(\alpha) = 0$.

Since fields are also integral domains, we have $g(\alpha) = 0$ or $h(\alpha) = 0$.

This contradicts the minimality of the degree of $f$. Thus minimal polynomials are irreducible.

Answer 1

Divide $f(x)$ by $p(x)$; you will get $f(x)=p(x)q(x)+r(x)$ with $\deg r(x)<\deg p(x)$. On the other hand$$0=f(\alpha)=p(\alpha)q(\alpha)+r(\alpha)=r(\alpha).$$But $p(x)$ is a minimal polynomial and $\deg r(x)<\deg p(x)$. So, $r(x)$ can only be the null polynomial, which means that $f(x)=p(x)q(x)$. So, $p(x)\mid f(x)$.

As you can see, the irreducibility of $p(x)$ is not needed at all.