Let $f(x) = x^4 + 3x^2 + 2x + 5$ in $\mathbb Q[x]$.
$f(x)$ has no rational roots.
$f$ is primitive and hence $f$ irreducible in $\mathbb Q[x] \iff f$ irreducible in $\mathbb Z[x]$.
Consider ring homomorphisms
$\mathbb Z[x] \to \mathbb Z_2[x]$
$\mathbb Z[x] \to \mathbb Z_5[x]$
Then we have that,
In $\mathbb{Z_2}[x]$: $\overline{f} = x^4 + x^2 + 1 = (x^2 + x + 1)^2$
In $\mathbb{Z_5}[x]$: $\overline{f} = x^4 + 3x^2 + 3 = x(x^3 + 3x + 2)$
If $f$ had been reducible in $\mathbb Z[x]$ then $f(x) = h_1(x)\cdot h_2(x)$
where $1 \leq \deg(h_1), \deg(h_2) \le 3$
Then $\overline{f(x)} = \overline{h_1(x)}\cdot \overline{h_2(x)}$ in $\mathbb Z_2[x]$ and $\mathbb Z_5[x]$.
This is a contradiction as we can't get the same degrees in $\mathbb{Z_2}[x]$ and $\mathbb Z_5[x]$.
Question
OK, I see that in $\mathbb Z_2[x]$ and $\mathbb Z_5[x]$ the factors have different degrees, but I don't understand how that implies $f$ is irreducible in $\mathbb Z[x]$. What theorem/lemma is that claim based on?
It's just reducing the factorization modulo $2$ and $5$. Consider cases: if $h_1$ has degree $2$, then reducing modulo $2$ gives no contradiction, but reducing modulo $5$ does. If $h_1$ has degree $1$ or $3$, then reducing modulo $5$ does not give a contradiction, but reducing modulo $2$ does.