Q:

Eisenstein's Criterion stronger version

So we all know about Eisenstein's Criterion and how useful it turns out to be. In particular, if we have a monic cubic $x^3+ax^2+bx+c$, it is irreducible if there exists a prime number $p$ such that $p\mid a,b,c$ and $p^2\nmid c$. 
But suppose that a monic cubic is irreducible, and all its coefficients are positive, but there does not exist such a prime. Sometimes we can find an integer $k$ such that we can substitute $x=y+k$ into the cubic, and the resulting polynomial in $y$ will satisfy the criterion. For example, if we have the cubic $x^3+3x^2+5x+1$, then no $p$ exists, but if we substitute $x=y+3$ into it, we end up with $y^3+12y^2+50y+70$, which, taking $p=2$, satisfies the criterion, hence it is irreducible. My question is whether or not this stronger version of the criterion applies to every irreducible polynomial. That is to say, if $x^3+ax^2+bx+c$ is irreducible, then we can always find some integer $k$ such that $(y+k)^3+a(y+k)^2+b(y+k)+c$, when simplified, will satisfy the criterion. 
Thank you all in advance!

A:

Take $f(x) = x^3 + x + 1\in \mathbb{Z}[x]$. It's irreducible, and 
$$f(x + k) = x^3 + 3k x^2 + (3k^2 + 1)x + k^3 + k + 1$$
has relatively prime linear and quadratic coefficients for any integer $k\not = 0$.

