Is $x^4+nx+1$ irreducible? 
Consider the polynomial $\xi= x^4+nx+1\in \mathbb Z[x]$. Show that if $n=\pm2$ then $\xi$ is reducible and that $n\neq\pm2$ implies $\xi$ is irreducible.

I got the answer by writing the possibilities out i.e. showing that $\xi=(ax^2+bx+c)(dx^2+ex+f)$ doesn't hold for any $n\in\mathbb Z$ and that $\xi=(ax+b)(cx^3+dx^2+ex+f)$ only holds for $n=\pm2$.
Though correct I'm looking for a more elegant answer. Can anyone help me find one?
 A: Not sure how much we can compress this :-(
The rational root test implies easily enough that the polynomial has a linear factor, iff $n=\pm2$.
That leaves the possibility of two quadratic factors with integer coefficients (Gauss' Lemma):
$$
x^4+nx+1=(x^2+ax+b)(x^2+cx+d).
$$
The constant term of the product is $bd=1$, so $b=d=\pm1$. But
$$
(x^2+ax+b)(x^2+cx+b)=x^4+(a+c)x^3+*+b(a+c)x+1,
$$
so in their product the coefficients of $x^3$ and $x$ terms, i.e. $a+c$ and $b(a+c)$, can only differ in sign. This is not the case with $\xi$, so such a factorization cannot exist.
A: $$(x^2+bx+c)(x^2+ex+f)=x^4+(e+b)x^3+(c+f+eb)x^2+(fb+ce)x+cf  
\equiv x^4+nx+1$$
So we shoud have: $$cf=1\,\,,\,e+b=c+f+eb=fb+ce=0$$  If $c=f=\pm 1$, $$c+f+eb=\pm2-b^2=0$$ Which hasn't any solution in $\mathbb Z$. So, this case, did not occur.
$$(x+b)(x^3+dx^2+ex+f)=x^4+(d+b)x^3+(bd+e)x^2+(be+f)x+bf\equiv x^4+nx+1$$
Which mean: $$bf=1\,\,,\,\,d+b=bd+e=0\,\,,\,\,\,be+f=n$$  If $\,b=f=1$ then: $\,d=-1\,\,,\,\,e=1\,\,,\,\,n=2$
If $\,b=f=-1\,$ then:  $\,\,\,d=1\,\,,\,\,e=-1\,\,,\,\,n=-2$
