$\text{Theorem: In an Integral Domain} \ R[x]$
$$\text{If}\ a \in U(R)\implies ax+b\ \text{is Irreducible in}\ R[x].$$
It's easy to prove it.
My question: Is the converse of this theorem right?
Example: Let be $f(x)=3x+2\in \mathbb{Z}[x]$. The element $$3\not \in U(\mathbb{Z})=\{+1,-1\} \kern.6em\not\kern -.6em \implies f(x)\ \text{is NOT Irreducible in}\ \mathbb{Z}[x].$$
My answer:
1st WAY: The only factorizations in $\mathbb{Z}[x]$ is $f(x)=1(3x+2)$ and $f(x)=(-1)(-3x-2)$ and that means that $f(x)$ is irreducile in $\mathbb{Z}[x]$.
2nd WAY: $f(x)$ is irreducible in $\mathbb{Q}[x]\iff f(x)$ is irreducible in $\mathbb{Z}[x]$ (because $f(x)$ is primitive).
Are all these things correct?