Is $x^{101} + 101x^{100} + 102$ irreducible?

Question

How can I show that $x^{101} + 101x^{100} + 102$ is irreducible over $\mathbb{Z}[x]$? I was not able to apply Eisenstein's criterion (Which was my first thought as 101 is prime) because of the $102$.

Answer 1

The irreducibility of $P(x)$ is equivalent to the irreducibility of $P(x − 1)$. Because the binomial coefficients $\binom {101}{k}$, $1 \leqslant k \leqslant 100$ are all divisible by 101, the polynomial $P(x − 1)$ has all coefficients except the first divisible by 101, while the last coefficient is $(-1)^{101} + 101(-1)^{101} + 102 = 202$, which is divisible by 101 (and not divisible by $101^{2}$). By Eisentstein's criterion, we have that $P(x-1)$ is irreducible, which implies that $P(x)$ is irreducible. Q.E.D.