$f(x)=2x^3+ax^2+bx+c$ where $a, b, c∈ℤ$. Prove that f is irreducible in $ℚ[x]$ if and only if $f({d\over2})≠0$ for all $d∈ℤ$.
Using a hint from my professor I have attempted a proof but I don't think I am proving it both ways for the iff...
A polynomial of degree 3 is irreducible iff it has no root.
Taking $f({x\over2})=2({x\over2})^3+a({x\over2})^2+b({x\over2})+c = {x^3\over4}+a{x^2\over4}+b{x\over2}+c $. Multiplying this by 4 we get a monic polynomial over $ℤ$: $x^3+ax^2+2bx+c$. Since $f({x\over2})$ is monic $cont(f)=gcd(a_0,...,a_n)=1$ so f is primitive. Since $f({x\over2})≠0$ it doesn't have a root therefore is irreducible over $ℤ$ which implies f is irreducible over $ℚ[x]$ (Gauss).
Note that $f(x)$ is irreducible iff $g(x) = 4f(\frac{x}{2})$ is irreducible.
Let's first assume that $f(\frac{x}{2})$ doesn't have any integer zeros. And you have already calculated that
$$g(x) = 4f(\frac{x}{2}) = x^3 + a x^2 + 2b + c.$$ The rational root theorem (not hard to prove) states that if $g(x)$ has a rational root $\frac{p}{q}$ then $p | c$ and $q | 1$. This implies in particular that $\frac{p}{q}$ would be an integer. By assumption $g = 4 f(\frac{x}{2})$ doesn't have any integer roots. So $g$ doesn't have any roots in $\mathbb{Q}$ and hence also $f$ doesn't.
Conversely, if $f$ is irreducible then it doesn't have any roots in $\mathbb{Q}$, which contains $\frac{1}{2} \mathbb{Z}$. So $f(\frac{d}{2}) \neq 0$ for all $d \in \mathbb{Z}$.