Let $f(x) = 2x^3+ax^2+bx+c$ where $a,b,c \in \Bbb Z$. Prove that $f$ is irreducible in $\Bbb Q[x]$ if and only if $f(d/2)$ does not equal $0$ for all $d \in \Bbb Z$.
I'm not really sure how to start this one. We know that a polynomial of degree $3$ is irreducible if and only if it does not have a root. I was thinking maybe to divide through by $2$ and then factorise but again not sure on how this would work. Any help would be appreciated. Thanks.
Continuing from your idea. Suppose that $m/n \in \Bbb Q$ is a root with $m \in \Bbb Z,\; n \in \Bbb Z^+, \; (m, n) = 1.$
Then, $n^3f(m/n) = 0$ or $$2m^3 + axm^2n + bmn^2 + cn^3 = 0.$$
Going modulo $n$ on both sides, we get: $$2m^3 \equiv 0 \mod n.$$
That is, $n$ divides $2m^3$. Now, since $m$ and $n$ are coprime, we also have that $(n, m^3) = 1.$
Thus, $n$ must divide $2$. In other words, $\dfrac{2m}{n} = \dfrac{2}{n}\cdot m \in \Bbb Z.$
To conclude, if $q \in \Bbb Q$ is a root of $f$, then $2q \in \Bbb Z$. Can you conclude now?