Yes, the four solutions to x^4+4=0 are x=±1±i. So to factor this completely, you have:
x^4+4=(x+1+i)(x+1-i)(x-1+i)(x-1-i)
If you multiply the first two factors and the second two factors, you obtain
x^4+4=(x^2+2x+2)(x^2–2x+2)
Alternatively, if you multiply the first and last factor, and also the second and third, you get:
x^4+4=(x^2–2i)(x^2+2i)
You can, of course, write the factorization in a third way, that being
x^4+4=(x^2+2ix-2)(x^2–2ix-2).