Assume i) $p|a_i \forall i<n$ ii) $p\not|a_n$ iii) $p^2\not|a_0$ Suppose $f(x)=g(x)h(x)$ where $g,h$ are polynomials with integer coefficients. Let $f(x)=a_nx^n+a_{n-1}x^{n-1}+.....+a_1x+a_0$, and let $g,h$ have coefficients $b_i, c_i$ similarly. $a_0 = b_0c_0$ implies, by (i) and (iii), that $p$ divides one of $b_0, c_0$. WLOG we assume $p$ divides $b_0$. By induction I show that, unless the degree of $g$ is $n$, $p$ divides every coefficient of $g$. Let $b_k$ be the first coefficient not divisible by $p$. $a_k = b_kc_0+b_{k-1}c_1+....+b_1c_{k-1}+b_0c_k$ (where some of the "c" coefficients may be zero). But if $k<n$, $p|a_k$ by (i), so $p|b_kc_0+b_{k-1}c_1+....+b_1c_{k-1}+b_0c_k$, but $p|b_{k-1}c_1+....+b_1c_{k-1}+b_0c_k$ by choice of $k$, so $p|b_kc_0$, and since $p$ does not divide $c_0$ (by above), it must divide $b_k$, a contradiction. Therefore, if the degree of $g$ less than $n$, $p|g$, so $p|a_n$, a contradiction of (ii), so the degree of $g$ must be $n$, whence the degree of $h$ is zero, and $f$ is therefore irreducible, which was to be shown.

That is irreducibilty in Z[x], now one more step to go ot Q[x].

Oh, I misread the question. Well, it's easy, since Gauss's Lemma states that a polynomial is irreducible over Z[x] iff it is irreducible over Q[x].