If p is a divisor of all coefficients of a polynomial f(x) but $p^2$ does not divide the constant. Show f(x) irreducible over $\mathbb{Q}$

Question

Let f = $r_0+r_1x+...+r_{n-1}x^{n-1} + x^n$. Assume that $p$ is a divisor of all coefficients but $p^2$ does not divide $r_0$. Show f(x) is irreducible over $\mathbb{Q}$

My attempt:

f = $r_0+r_1x+...+r_{n-1}x^{n-1} + x^n$ suppose f is reducible. Then there exists $g,h$ s.t. $f=gh$ and $0 < deg(g), deg(h)<deg(f)$. Then $p$ divides $f-x^n=gh-x^n$, but I get stuck here. I also have this theorem that says if f is irreducible over $\mathbb{Z}_p$ then its irreducible over $\mathbb{Q}$ so that may be useful here

Answer 1

We have to suppose that both $g(x)$ and $h(x)$ are divisible by $p$ and are of the form: $$g(x)=bx^r\pmod p$$ $$h(x)=cx^s\pmod p$$ $$f(x)=g(x)h(x)=ax^{r+s}\pmod p$$

That both the terms of grade $0$ of $g$ and $h$ are divisible by $p$.

But $f(x)=g(x)h(x)$ and the coefficient $r_0$ is divisible by $p^2$ (because every number is a divisor of zero) and this is a contradiction.

This means that $f(x)$ must be irreducible modulo $p$.

Answer 2

As you wrote, suppose $\;f=gh\;$ . But we have

$$f=x^n\pmod p=\left(g\pmod p\right)\cdot \left(h\pmod p\right)$$

Check now what happens with the free coefficient...