Can a polynomial be almost always divisible by a member of a finite set of primes? Special case of Bunyakovsky conjecture
Let $f(x)$ be non-constant irreducible polynomial with integer
coefficients, no fixed prime factor and positive
leading coefficient. Let $S$
be a finite set of primes.

Q1 Is it possible $f(n)$ to be divisible by a member of
  $S$ for almost all natural $n$?
Q2 Must $f(n)$ take at least one prime value for
  natural $n$?

 A: The answer to Q1 is "no". The value of $f(n)$ mod $p$ depends only on the value of $n$, mod $p$.
Let $N$ be the product of the primes in $S$. Assuming "almost all" means "all but finitely many", then that means that for $n$ sufficiently large, the set $n+1,\ldots,n+N$ are all multiples of primes from $S$. But since all this is invariant modulo $N$, that means every number is a multiple of some prime from $S$.
The question is then, can we cover $1,\ldots,N$ by residue classes mod primes from $S$ without using all the residue classes modulo any prime (because if we did we have a fixed prime factor)? And the answer is "no", by an easy application of the Chinese Remainder Theorem: choose a residue class modulo each $p\in S$ which is not used; they give a unique residue mod $N$ which is uncovered.
A: I guess my answer is morally the same as James Cranch's one, but Q1 is false even in degree $1$: let $f(x)=ax+b$ and $\{p_1,\dots,p_s\}=S$. Then by assumption $f$ is not the constant polynomial $0$ modulo any $p_i$ and we let $\alpha_i$ be an integer which is a non-root $\mod{p_i}$. The Chinese Remainder Theorem says that the finite set of equations
$$
\left\{
\begin{array}{l}
x\equiv \alpha_1 \pmod{p_1}\\
\vdots\\
x\equiv \alpha_s\pmod{p_s}
\end{array}
\right.
$$
has infinitely many solutions.
