$F\left(S\right)=\dfrac{3s}{\left(s+6\right)\left(s+2\right)}=\dfrac{A}{s+6}+\dfrac{B}{\left(s+2\right)}$ $Poles\ :s=6,s=-2$ $zeros\ :s=0$ $partial\ fraction\ method:\ \ 3s=A\left(s+2\right)+B\left(s+6\right)$ $when\ s=-2,\ \ \ s=-6$ $B=\dfrac{-3}{2};\ \ A=\dfrac{9}{2}$  Graphical method : $A=H\frac{Phasor\ for\ zero\ at\ origin\ to\ pole\ at\ Q}{Phasor\ from\ pole\ at\ Q\ to\ pole\ at\ P} =3\dfrac{6\angle{}108^\circ{}}{4\angle{}180^\circ{}}=\dfrac{9}{2} $  $B=H\frac{phasor\ from\ zero\ at\ origin\ to\ pole\ at\ P}{\begin{array}{l}phasor\ from\ pole\ at\ P\ to\ pole\ at\ Q \ \ \end{array}}$ $=3\dfrac{2\angle{}180^\circ{}}{4\angle{}0^\circ{}}=\dfrac{-3}{2}\ \ \left{\because{}\dfrac{3}{2}\angle{}180^\circ{}\right}$  $\therefore{}F\left(s\right)=\dfrac{9}{2\left(s+6\right)}\ \ \ \ \ \dfrac{-3}{2\left(s+2\right)}$ Taking inverse Laplace transform $F\left(t\right)=\dfrac{9}{2}\ e^{-6t}\dfrac{-3}{2}\ e^{-2t}$