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Given a polynomial $P(x)$ in $Q[x]$, can I always find a rational number, $c$, so that $P(x)+c$ is irreducible in $Q[x]$?

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    By clearing denominators, which does not affect whether a polynomial can be factored, we may as well assume $P(x) \in \Bbb Z[x]$, in which case a factorization over $\Bbb Q$ is necessarily a factorization over $\Bbb Z$. – Travis Willse May 24 '16 at 10:25
  • The answer is almost surely yes, but I have no idea for a proof yet. – Peter May 24 '16 at 10:30

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I just recalled a neat lemma which solves this: Like Travis said we first move to $Z[x]$ and use the following lemma:

If a polynomial $P=a_nx^n+..+a_0 $ over $Z$ satisfies $a_0$ is prime and $|a_0|>|a_n|+|a_{n-1}|..+|a_1| $ then it is irreducible. A proof can be found here: http://yufeizhao.com/olympiad/intpoly.pdf page 2