Factoring $z^4+64=0$

Question

I've been asked to factor $z^4+64=0$ in the structure $(z^2+Az+B)(z^2+Cz+D)=0$, where $A$, $B$, $C$ and $D$ are real numbers. I've found the four roots of the equation, but I don't know how to find those structures. Should I try combining the roots until I get that structure or is there any less 'artificial' method? I thought about $(a+b)(a+b)=(a^2+2ab+b^2)$, but there are some i I can't avoid. Thank you.

Answer 1

$$ (z^2+Az+B)(z^2+Cz+D)=z^4+(A+C)z^3+(AC+B+D)z^2+(AD+BC)z+BD $$ so we get a system of equations in $A,B,C,D$ $$ A+C=0 \\ AC+B+D=0 \\ AD+BC=0 \\ BD=64 $$ If you solve it, you get $A=4,C=-4,B=D=8$ and thus $z^4+64=(z^2+4z+8)(z^2-4z+8)$.

Answer 2

$$z^4+64=z^4+16z^2+64-16z^2=(z^2+8)^2-(4z)^2=(z^2+4z+8)(z^2-4z+8)$$