Is the polynomial $(2x + 2)$ irreducible in $\mathbb{Z}[x]$?
I know that the units in $\mathbb{Z}[x]$ are $\pm$1.
Please check my proof below:
Suppose $f$ = $(2x + 2)$ $\in$ $\mathbb{Z}$[x] is a degree 1-polynomial. Then $f=1(2x + 2)$ where $g = 1$ and $h = (2x + 2)$ for $g, h$ $\in$ $\mathbb{Z}$[$x$]. Since $g$ is a unit, we have $f =\text{unit}\cdot$degree-$1$ polynomial. Hence $f$ is irreducible in $\mathbb{Z}[x]$ .
On
$(2x+2)$ is reducible in $\mathbb{Z}[x]$.
To see why let's start with the definition.
An element $f$ in a commutative ring $R$ is called irreducible if there does not exist non-units(Or non-invertible elements) $x$ and $y$ such that $f = xy$.
Thus an element is reducible if it can be written as the product of two non units.
We can write $2x+2$ as $2 * (x+1)$, both of which are non-units in $\mathbb{Z}[x]$.(The only units in $\mathbb{Z}[x]$ are $1$ and $-1$). Thus $2x+2$ is reducible.
Your proof is incorrect - you display one factorisation in which one of the factors is a unit. However to be irreducible you have to consider all possible factorisations.
As in the comments the factorisation $2x+2=2(x+1)$ is a factorisation into non-units in $\mathbb Z[x]$ which shows that the polynomial is reducible.
The situation in $\mathbb Q[x]$ is different, because $2$ is a unit in this context.
You don't need Eisenstein here - since the product of polynomials has degree the sum of the degrees, and a linear polynomial has degree $1$ the only possible factors have degree $0$ (constants) or $1$. If there is a factorisation in the $\mathbb Z$ context, one of the factors must be a constant, and this will be a product of primes. The prime factors of the constant must clearly be factors of each of the coefficients of the polynomial, and any prime which divides the coefficients leads to a factorisation.
In the $\mathbb Q$ context every constant other than $0$ is a unit, so a non-trivial factorisation into irreducibles must involve two factors of degree greater than $0$.