Solution:

Foster −1 from:

$ \begin{aligned} \\ &z(s)=\frac{3(s+2)(s+4)}{(s)(s+3)}....(1)\\\\ &=\frac{3 s^2+18 s+24}{s^2+3 s}\\\\ &\therefore z(s)=3+\frac{9 S+24}{s^2+3 S}\\\\ &=Z_1(s)+Z_2(s) ...(2)\\ \end{aligned} \\ $

Where

$ \quad z_1(s)=3 \\ $

And

$ \left.\begin{array}{rl} \\ z_2(s) & =\frac{9 s+24}{(s)(s+3)} \\\\ & =\frac{A}{s}+\frac{B}{s+3} \\ \end{array}\right\}....(3) \\ $

$ \begin{aligned} \\ &A=\left.\frac{9 S+24}{(s)(S+3)}\right|_{S=0}=\frac{0+24}{0+3}=8 \\\\ &B=\left.\frac{9 S+24}{S}\right|_{S=-3}=\frac{-27+24}{-3}=1 …