Showing that the polynomial is irreducible Let $a \in 3 + 4\mathbb{Z}$ and $f\left( x \right) = \left( {x + 1} \right)\left( {x - 1} \right)\left( {x + 4} \right)\left( {x - 2} \right) + a$ a polynomial. Using the homomorphism $\pi :\mathbb{Z} \to \mathbb{Z}/4\mathbb{Z}$, see if $f$ is irreducible over $\mathbb{Q}$.
My attempt: It's easy to show that all rational zeroes of $f$ must be integers and that there are no integer zeroes. Also, I can show that if $\pi \left( f \right)\left( x \right) = \overline 1 {x^4} + \overline 2 {x^3} + \overline 3 {x^2} + \overline 2 x + \overline 3 $ is irreducible, then so is the $f$ (notation $\overline k  = k + 4\mathbb{Z}$). However, the Eisenstein criterion fails since the only ideals in $\mathbb{Z}/4\mathbb{Z}$ are $\mathbb{Z}/4\mathbb{Z}$, $\left\{ {\overline 0 ,\overline 2 } \right\}$ and $\left\{ {\overline 0 } \right\}$, of which $\left\{ {\overline 0 ,\overline 2 } \right\}$ and $\left\{ {\overline 0 } \right\}$ are prime, and $\overline 3 $ is not contained in any of the prime ideals.

Modulo $2$, we have
$$(x^2+x+1)^2 = x^4 + 2x^3+3x^2+2x+3$$
so any monic factor of $ x^4 + 2x^3+3x^2+2x+3$ in $\mathbf Z/4\mathbf Z$ reduces to $x^2+x+1$ modulo $2$. There are only $4$ possible factors to check.
