I am asked to show that $x^3-nx+2$ is irreducible if $n\neq -1, 3, 5$.
I proved a similar problem before, where I was asked to show that the polynomial $f(x)=x^3+nx-2$ is irreducible over $\mathbb{Z}$ if $n\neq 1, -3, -5$. (Proof below)
$Proof:$ Let $p(x)=x^3+nx+2$ in $\mathbb{Z}[x]$, and suppose that $p(x)$ is reducible. By the rational root test, a root $a$ of $p(x)$ must be $a=\pm1, \pm 2$.
If $p(1)=0$, then $1^3+n(1)+2=0\implies n=-3$.
If $p(-1)=0$, then $(-1)^3-n+2=0\implies n=1$.
If $p(2)=0$, then $(2)^3+2n+2=0\implies n=-5$.
If $p(-2)=0$, then $(-2)^3-2n+2\implies n=-3$.
Therefore, if $n\neq1,-3,-5$, then $p(x)$ is irreducible.
My question is, how should I go about this proof differently than the previous one, since I do not know what field I am factoring over? (i.e. I do not think I can use the rational root test because we are not necessarily factoring over $\mathbb{Z}$ or $\mathbb{Q}$?)
This question is from Dummit and Foote's algebra book, section 13.1 question 7 by the way.