How to factor $n$-degree polynomials of this form?

Question

I have come across a very specific form of polynomial, and I was hoping there would be a nice way to factor it - or at least show that it is irreducible. The form is $x^{n+1} + 2k \cdot x^n -1$ where $k$ is some fixed integer. I'm not very familiar with abstract algebra or complex analysis, which I feel like would be necessary in factoring this polynomial. Can it be done?

Answer 1

If $k=0$ it's always reducible : $x^{n+1}-1=(x-1)(x^n+...)$
If $k=1$ and $n$ is an even number, then it's reducible, because $-1$ is a root of $x^{n+1}+2x^n-1$

For $k\neq 0,1$ and $k=1$ when $n $ is an odd number, it seems irreducible. But I think it's difficult to show. Because we can't apply Eisenstein's criterion to $f(x)=x^{n+1}+2kx^n-1$ or apply this for $f(x+m)$ for some $m\in \mathbb Z$.
For $k=-1,\pm2,\pm3,\pm4,\pm5$ I checked it for $n\leq 100$ by Mathematica, and in all cases they are irreducible.