Browse · MATH Print → jmc algebra intermediate Problem Expand (z2−3z+2)(z3+4z−2). Solution — click to reveal ×\cline1−7+\cline1−7z5z5−3z4−3z4z3+2z3+4z3+6z3z2−12z2−2z2−14z2+4z−3z+8z+6z+14z−2+2−4−4So, our answer is z5−3z4+6z3−14z2+14z−4. Final answer z^5-3z^4+6z^3-14z^2+14z-4 ← Previous problem Next problem →