Question

How do you do this one?

Answer 1

Your answer is correct... $i^2=-1$, so

$(3+i)(i^2+8i-2)=(3+i)(-3+8i)$

Then distributing $3+i$ to both terms on the right gives

$(3+i)(-3)+(3+i)(8i)=(-9-3i)+(24i+8i^2)=-9-3i+24i-8=-17+21i$