Is this polynomial irreducible?

Question

Let $n \in \mathbb{N}^*$ and $p$ be a prime number. Is the polynomial $f=X^{2n+1}+(p+1)X^{2n}+p \in \mathbb{Z}[X]$ irreducible?

Answer 1

I once remarked in an answer to Show an infinite family of polynomials is irreducible that even in the "equality case" one can use Perron, given that the polynomial $P(z)$ has no zero on the unit circle.

For this you need to use the extended Rouche's theorem (the "Symmetric Version" in https://en.wikipedia.org/wiki/Rouch%C3%A9%27s_theorem ).

Indeed, take $f = P(z)$ and $g = a_{n-1} z^{n-1}$. Then $|f(z)| + |g(z)| > |g(z)| = |a_{n-1}| \ge 1 + |a_{n-2}| + ... + |a_0| \ge |f(z) - g(z)| $ on $|z| = 1$.

So you know that $P(z)$ has $n-1$ roots (strictly) inside the unit circle. Proceed as in the article you quoted from now on to show that $P(z)$ is irreducible.

Apply this here.