Irreducible polynomial over Q Let $f(x) = 3x^4+6x^3+24x^2+18 \in \mathbb Z[x]$. Is $f(x)$ irreducible over $\mathbb Q$ ?

In my course, Eisenstein's criterion is apply for monic polynomial only, hence, I can't use it with p =2. If I can, everything turn easy. 

My attempt: If $f(x)$ is irreducible over $\mathbb Z$, then $f(x+1)$ is irreducible as well. Replace x by x+1, I have $f(x+1) = 3(x+1)^4+6(x+1)^3+24(x+1)^2+18\\=3x^4+18x^3+60x^2+78x+51$ 

But in mod 3, I have f(x+1)=0 I don't know how to argue further. Please, explain me
Since $3$ is invertible in $\mathbb Q$, $f$ is irreducible in $\mathbb Q[X]$ iff so is $\frac f3=x^4+2x^3+8x^2+6$. Now you can apply Eisenstein's criterion.
I have given two speeches in South America over a decade ago with an irreducibility criteria I established: “if a polynomial $h(x)$ of $\mathbb{Z}[X]$ has degree $n$ and $h(n)$ is prime for $2n + 1$ rational integers then $h$ is irreducible”. I apply this result here. The nine couples $(n, h(n)) = (1, 17), (-1, 13), (-7, 2113),$ $(-11, 12953), (13, 34313), (19, 146933),$ $(31, 990797), (43, 3592613), (-65, 17335181)$ where the nine $h(n)$ are prime, prove, according to this result, that $h = \frac{f}{3}$ is irreducible over $\mathbb{Z}$ an so is $f$.

NOTE: This criterion is not very comfortable to apply but could be useful sometimes in algebraic number theory.