Evaluate the integral

${\displaystyle \underset{1}{\overset{2}{}}}(-24z^{3}4z)dz$

given

${\displaystyle \underset{1}{\overset{2}{}}}{x}^{3}dx=\frac{15}{4}$

${\displaystyle \underset{1}{\overset{2}{}}}{x}^{2}dx=\frac{7}{3}$

${\displaystyle \underset{1}{\overset{2}{}}}xdx=\frac{3}{2}$

${\displaystyle \underset{1}{\overset{2}{}}}dx=1$