Let \(f\in \ZZ[x]\) be monic. If there exists a prime \(p\) such that the reduction \(\bar f\in\FF_p[x]\) is irreducible, then \(f\) is irreducible in \(\QQ[x]\).
Suppose \(f=gh\) with \(g,h\in\QQ[x]\) of degree at least \(1\). By Gauss’s lemma we may take \(g,h\) monic in \(\ZZ[x]\). Reducing modulo \(p\) gives \(\bar f=\bar g\bar h\), and since \(g\) and \(h\) are monic, \(\deg\bar g=\deg g\geq1\) and \(\deg\bar h=\deg h\geq1\). This contradicts the irreducibility of \(\bar f\).
The converse fails: \(x^4+1\) is irreducible in \(\QQ[x]\) but reducible modulo every prime \(p\). Modulo \(2\), \(x^4+1=(x+1)^4\). For \(p\) odd, at least one of \(-1\), \(2\), \(-2\) is a square modulo \(p\), since the product of two nonsquares is a square. If \(i^2=-1\) then \(x^4+1=(x^2-i)(x^2+i)\); if \(c^2=2\) then \(x^4+1=(x^2+cx+1)(x^2-cx+1)\); if \(c^2=-2\) then \(x^4+1=(x^2+cx-1)(x^2-cx-1)\). For a monic irreducible \(f\in\ZZ[x]\), the Chebotarev density theorem shows that \(\bar f\) is reducible for every prime \(p\) exactly when the Galois group of \(f\) over \(\QQ\) contains no \((\deg f)\)-cycle; the Galois group of \(x^4+1\) over \(\QQ\) is \((\ZZ/2\ZZ)^2\).