Avoiding primefactors in reducible polynomials Take distinct pairs $(c_i,d_i) \in \mathbb Z^2$, the entries being coprime. Put $f(x) = \prod_{i=1}^k (c_i x + d_i)$. Let $\mathbb P$ denote the rational prime numbers. Which conditions (if any) need to be imposed on $f$ in order to satisfy
$$\forall \textrm{ finite sets } S \subseteq \mathbb P: \exists n \in \mathbb N : \forall p \in S :\gcd(f(n),p)=1$$
If $k=1$ Dirichlet's theorem on primes in arithmetic progressions does the job.
 A: One need only test the condition for single primes because $f(x)\pmod p$ depends only on $x\pmod p$; then multiple primes can be combined using the Chinese remainder theorem.
Let $p$ be a prime. 
A factor $c_ix+d_i$ is a multiple of $p$ for at most one residue mod $p$ and we can ignore the factors with $c_i\equiv 0\pmod p$.
From all other $i$, we get the set $c_i^{-1}d\pmod p$. If and only if this set does not cover all of $\mathbb Z/p\mathbb Z$, a solution of $f(x)\not\equiv 0\pmod p$ exists. This is trivially the case for all $p>k$ but needs to be tested for all smaller $p$. A simple criterion is that $\gcd(f(0),f(1),\ldots,f(k-1))$ must have only prime divisors $> k$.
For example, with  $f(x)=(x+5)(2x+3)(3x-1) $ we have $\gcd(f(0),f(1),f(2))=5$, which is ok.
A: This is not an answer or exactly what you are looking for. I see Hagen von Etizen has already given a pretty good answer, I am here providing some basic notion of Sieves regarding this problm as you have asked.
Let, $N:=N(S)=\prod\limits_{p\in S}p$ be a square free number. Your goal is to find $f(x) = \prod_{i=1}^k (c_i x + d_i)$ such that $g(n)=\sum\limits_{d|(N,f(n))}\mu(d)$ is not a zero function.
You can also formulate as, $G(x)=\sum\limits_{n\leq x}g(n)$ is positive anywhere or not. You may clearly note that if contrary of your condition "$\forall \textrm{ finite sets } S \subseteq \mathbb P: \exists n \in \mathbb N : \forall p \in S :\gcd(f(n),p)=1$" happens if and only if $G(x)$ is a zero function.
See, $G(x)=\sum\limits_{n\leq x}\sum\limits_{d|f(n),d|N}\mu(d)=\sum\limits_{d|N}\mu(d)t_d(x)$, where $t_d(x)=\#\{f(n):d|f(n), n\leq x\}$.
Now, formal procedure of Sieve theory plans to have, $t_d(x)=xg(d)+r(d)$ where $g(n)$ is absolutely multiplicative.
$t_d(x)=\#\{f(n):d|f(n), n\leq x\}=\sum\limits_{m\in \mathbb{Z/d\mathbb{Z}}}\#\{n\leq x: d|f(m), n \equiv m (\text{mod}\ d)\}$.
Thus, take $g(d)=\#\{m(mod\ d):d|f(n)\}$. Using this, you could have $G(x)=x\sum\limits_{d|N}\mu(d)g(d)+\text{error term}$.
It can be shown that $G(x)$ ~ $x\prod\limits_{p\in S}(1-g(p))$
