## Calculus with Applications (10th Edition)

$\displaystyle \frac{1}{2}$ and $-4$
Set the $RHS$ to 0 by subtracting 4 from both sides $z(2z+7)=4$ $\qquad.../-2$ $2z^{2}+7z-4=0$ To factor the trinomial, we search for factors of $2\times(-4)=-8,$ whose sum is $+7:\qquad$ We find $+8$ and $-1$. $2z^{2}+7z-4=2z^{2}-z+8z-4=$ ... and factor in pairs: $=z(2z-1)+4(2z-1)$ $=(2z-1)(z+4)$ $(2z-1)(z+4)=0$ By the zero product principle, $2z-1=0$ or $z+4=0$ $z=\displaystyle \frac{1}{2}$ or $z=-4$ The solutions are $\displaystyle \frac{1}{2}$ and $-4$.