Is $x^4+4$ an irreducible polynomial?

We know that $p(x)=x^4-4=(x^2-2)(x^2+2)$ is reducible over $\mathbb{Q}$ even not having roots there.

What about $q(x)=x^4+4\in \mathbb{Q}[x]$? Again, no roots.

$$\begin{eqnarray}x^4+4&=&(x^2+2i)\cdot (x^2-2i)\\ &=& (x-(1-i))\cdot (x+(1-i))\cdot (x-(1+i))\cdot(x+(1+i)) \\ &=& ((x-1)+i)\cdot ((x-1)-i)\cdot((x+1)-i)\cdot((x+1)+i) \\ &=& ((x-1)^2+1)\cdot((x+1)^2+1).\end{eqnarray}$$

Reducible.

$x^4+4 \cdot 1^4= x^4+ 2 \cdot 2 \cdot x^2+2^2 - (2x)^2$

Which is well known identity called Sophie Germain

As Berci showed, this polynomial is indeed reducible over the rationals. One way to see it is to calculate its roots explicitly: $$x^4+4=0 \leftrightarrow x^2 = \pm 2i \leftrightarrow x = \pm \sqrt{2} (\frac{\sqrt{2}}{2}(1+i)) \vee x = \pm i\sqrt{2} (\frac{\sqrt{2}}{2}(1+i)) $$ Or: $$x = \pm 1 \pm i$$ And since those roots are proper complex number in $\mathbb{Z}[i]$, you can pair $1+i$ with $\overline{1+i}=1-i$ and $-1+i$ with $\overline{-1+i} = -1-i$ and obtain the factorization $(x^2 - 2x + 2)(x^2 + 2x +2)$ (if $\alpha$ is a proper complex root of $p \in \mathbb{R}[x]$, then $\overline{\alpha}$ is another root, and $(x-\alpha)(x-\overline{\alpha}) = (x^2-2Re(\alpha) + |\alpha|^2)$ divides $p$.

$$X^4+4=X^4+4X^2+4-4X^2 =(X^2+2)^2-(2X)^2 \,.$$