Precalculus (6th Edition)

$\color{blue}{2z-z^{3/2} -z^2}$
RECALL: $(a+b)(c+d)=ac + ad + bc + bd$ (more popularly known as the FOIL method. Use the formula above to obtain: $=2z^{1/2}\cdot z^{1/2} + 2z^{1/2} \cdot (-z) + z \cdot z^{1/2} + z \cdot (-z)$ Use the rule $a^m \cdot a^n = a^{m+n}$ to obtain: $=2z^{1/2+1/2} -2z^{1/2+1}+z^{1+1/2}-z^{1+1} \\=2z^1-2z^{3/2} + z^{3/2} -z^2 \\=\color{blue}{2z-z^{3/2} -z^2}$