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. 

One may use the same version of completing the square that proves that $x+\dfrac1x \ge 2$ when $x>0$:
$$
x+\frac1x = \left(x-2+\frac1x\right)+2 = \left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^2+2.
$$
Similarly
$$
x^4+4 = \left( x^4 +4x^2 + 4 \right) - 4x^2 = \left(x^2+2\right)^2 - (2x)^2
$$
then factor that as a difference of two squares.
