Let $\displaystyle f(x)=a_0x^n+a_1x^{n-1}+...+a_{n-1}x+a_n $be a polynomial with integer coefficients and assume that for a given prime integer $\displaystyle p $that:

1. $\displaystyle p $ does not divide $\displaystyle a_0$,

2. $\displaystyle p $ divides $\displaystyle a_1,a_2,...,a_n$

3. $\displaystyle p^2$ does not divide $\displaystyle a_{n-1}$

Prove that either$\displaystyle f(x)$ has a rational root, or$\displaystyle f(x) $is irreducible over the filed $\displaystyle Q$

What is the meaning of the last sentence? Do I have to show that there are two cases: case1 irreducible, case2 has rational root?

I assume that I should consider two cases. Case1 is easy, it occurs when $\displaystyle p^2$ does not divide $\displaystyle a_{n}$. In this case Eisenstein's criterion gives answer. But what about case2 when $\displaystyle p^2$ divides $\displaystyle a_{n}$ ?

I managed to show the following in case2:

$\displaystyle f(x)=x(a_0x^{n-1}+a_1x^{n-2}+...+a_{n-1})+a_n$ where $\displaystyle a_0x^{n-1}+a_1x^{n-2}+...+a_{n-1}$ is irreducible according to Eisenstein's criterion. I have a feeling that $\displaystyle f(x) $has a rational root, but do not know how to show it...