Factorise $p(z)=z^5+z^4-8z^3-28z^2+16z+96$ into linear factors.

 The original question was that given $-2-2i$ is a root and that $3$ other roots were real, factorise into linear factors. Since $p(z)$ has real coefficients this means the conjugate root is also a root, so both $-2-2i$ and $-2+2i$ are roots. 

Here we assume that there are $3$ real roots and go from there.