Consider the following family of polynomials for every integer $d > 0$:
$$P_d(X) = X^{d+1} – X^d – 1$$
I was wondering if these were irreducible (over $\mathbb{Q}$) or not. Checking the first few hundred values of $d$ with Mathematica suggests that $P_d(X)$ is reducible iff $d \equiv 4\ (\text{mod}\ 6)$. I checked all the irreducibility criteria I know but didn't find anything that works.
Some more information on the roots:
These polynomials each have one positive real root $x_0\in(1,2)$ and, if $d$ is odd, one negative real root on $(-1,0)$. All other roots are non-real with modulus $<x_0$. The polynomial can easily be shown to be squarefree, so all roots are distinct.
Possible generalisation:
I also checked for constant terms $a_0$ other than $-1$. For $a_0 = 1$, it seems that it is reducible iff $d > 1$ and $d \equiv 1\ (\text{mod}\ 6)$. For $a_0 = 2$ it seems to be reducible for all even $d$, and if $a_0 = -2$ for all odd $d$. For $a_0$ some other non-zero integer, almost all of them seem to be irreducible except sporadically (e.g. for $a_0 = -6$ and $d = 1$ it factors and for $a_0=-4$ and $d = 2$). I mostly care about the case $a_0 = -1$ though, so I haven't thought much about these.
Best Answer
The irreducibility is equivalent to that of $−X^{d+1}P_d(1/X)=X^{d+1}+X−1$, and method from Keith Conrad's answer in Irreducibility of $x^n-x-1$ over $\mathbb Q$ should lead to the proof. Especially look at the lecture notes refered in the comments there, it contains your case as a theorem.
Also one more reference, in book Polynomials by Victor V. Prasolov, see section 2.3.2 Irreducibility of certain trinomials, the irreducibility of trinomials of form $x^n \pm x^m \pm 1$ is discussed.