Let $f(x)$ be a polynomial over $\mathbb{Z}$. How to show that there is infinite number of primes $p$ such that $f(x)$ splits into linear factors over $\mathbb{F}_{p}$? Particularly, the task I was given is to solve this problem using the fact that $\lim_{s\rightarrow 1}{(s-1)\zeta_{K}(s)}$ is finite and nonzero. ($K$ is an extension of $\mathbb{Q}$, generated by all the roots of $f(x)$.)
Here is what I have understood so far: all primes might be divided in two groups, namely the first group consists of primes that ramify in K, and the second, of course, consists of primes that do not. Since only finite number of primes ramify in finite extensions, and since every prime number decomposes uniquely into a finite product of prime ideals of the ring of integers, the part of Dedekind zeta-function of K corresponding to the ramified primes converges for s=1. The second group of primes then must be infinite. Further we can divide this second group into primes that do not have a prime ideal of inertia degree one in their decomposition, and primes that do. Similarly, the number of primes that do have such an ideal in their decomposition must be infinite. These primes contain ones that I need, but I do not know how to proceed from this point. Could anyone help me, please?