Let $f(x)=c_nx^n+c_{n-1}x^{n-1}+\cdots+c_1x+c_0$ be a polynomial in $\mathbb{Z}[x]$ such that for some prime $p$ we have $p$ does not divide $c_n$, $p$ does not divide $c_{n-1}$, $p$ divides each other $c_i$, and $p^2$ does not divide $c_0$. Show that $f$ is irreducible in $\mathbb{Q}$ if and only if there does not exist a rational number $a/b$ such that $f(a/b)=0$.
The $\Rightarrow$ direction is straightforward. I'm not sure how to tackle the $\Leftarrow$ direction. Clearly, the polynomial $g(x)=c_{n-1}x^{n-1}+\cdots+c_1x+c_0$ is irreducible by Eisenstein's criterions using $p$. However, extending this seems to be a challenge. I've played around with plugging in $a/b$ to show that $f(a/b)\neq0$ is irreducible, but didn't get far. Also tried an inductive argument on the degree of the polynomial, but didn't see how to argue for degree above 3. I've considered the contrapositive statement, but I don't know of any necessary conditions for reducibility that imply a rational root. If there are any suggestions on how to approach this, that would be greatly appreciated. Full solution not required.
Suppose that $f$ is reducible.
Then,
$c_n x^n+c_{n-1}x^{n-1}+...+c_1x+c_0=(d_s x^s+...+d_1x+d_0)(e_tx^t+...+e_1x+e_0)$
Without loss of generality, suppose $p\not\mid e_0$. Let $i>0$ be the minimum integer such that $p\not\mid d_i$ such an $i$ must exist as otherwise $p$ divides all of the $d_j$ and then, $p$ divides all of $a_j$.
Claim: $p\not \mid a_i$ and $p \mid a_j$ for every $j<i$.
Proof of Claim: $a_i=d_i e_0+d_{i-1} e_1+...+d_0 e_i$ and $p\mid d_{i-1} e_1+...+d_0 e_i$ but $p\not\mid d_i e_0$. Hence, $p\not\mid a_i$. For $j<i$, $a_j=d_je_0+...d_0e_j$ so $p\mid d_j$.
So we know that $i=n-1$ as $p$ divides $a_j$ for all $j<n-1$. In particular, we learn that $s\geqslant n-1$. So, $e_tx^t+...e_1x_1+e_0$ is a constant factor if $s=n$ or a linear factor if $s=n-1$. (The former case can be ignored as that does not give us a non-trivial factorization.) So