Is $x$ irreducible in $(\mathbb Z/{12\mathbb Z})[x]$? I have a couple of doubts regarding irreducibility of $x$ over $Z/{12Z}$.
First is w.r.t definition of irreducibility : 
When we talk of a polynomial $p(x)$ being irreducible in $Z_{n}$, we basically mean to say that $p(x)=0$ has no solutions in $Z_{n}$ right?
If so then $x$ is not irreducible in $Z_{n}$.
I arrived at this result by showing that $A=(x)$ the ideal generated by $x$ is not maximal by proving that $Z_{12}[x]/A$ is not a field.
My attempt is as follows:
The elements of $Z_{12}[x]/(x)$ are of the form $c+A$ where $c \in Z_{12}$. But since $Z_{12}$ is not a field, $c^{-1}$ might not exist. Hence $Z_{12}[x]/(x)$ is not a field thus $x$ is not irreducible over $Z_{12}$.
Could someone check whether my attempts are right?
EDIT:
I know that a polynomial $p\in R[x]$ is said to be irreducible if it cannot be factored into the product of polynomials $f,g \in R[x]$ such that one of them is a constant.

Irreducible means the following: we say that $r\in R $ is irreducible if $r = ab $ then either $a$ or $b$ is a unit. So in $\mathbb Z_{12}$, for example, $2$ is irreducible but $3$ is not since $3 = 3(-3) \mod 12 $ in general we have $p\mid n $ is reducible iff $p^2 \nmid n$. 
So let's return to our ring $\mathbb{Z}_{12}[x]$ and consider $x=(3x+4 )(4x+3)= 12x^2 + 25x + 12 \bmod 12 $ hence $x$ is reducible (if we proved the factors are non units) as you said. Now we need to show that $3x+4$ and $4x+3$ are not units in $\mathbb{Z}_{12} [x]$. 
You can  show that $3x+4$ is a not a unit by contradiction. 
