I suppose you're asking about the inverse Laplace transform of
[tex]F(s)=\dfrac4{(s-3)(s-2)}[/tex]
by way of the convolution theorem.
The theorem itself says that
[tex]\mathscr L^{-1}\{A(s)B(s)\}=(a*b)(t)[/tex]
where [tex]A(s)[/tex] and [tex]B(s)[/tex] are the Laplace transforms of [tex]a(t)[/tex] and [tex]b(t)[/tex], respectively.
Let [tex]A(s)=\frac1{s-3}[/tex] and [tex]B(s)=\frac1{s-2}[/tex]. Then
[tex]a(t)=\mathscr L^{-1}\left\{\dfrac1{s-3}\right\}=e^{3t}[/tex]
[tex]b(t)=\mathscr L^{-1}\left\{\dfrac1{s-2}\right\}=e^{2t}[/tex]
So we have
[tex]\mathscr L^{-1}\left\{\dfrac4{(s-3)(s-2)}\right\}=e^{3t}*e^{2t}[/tex]
Convolution is defined as
[tex](a*b)(t)=\displaystyle\int_0^t a(t-\tau)b(\tau)\,\mathrm d\tau[/tex]
The convolution of [tex]e^{3t}[/tex] and [tex]e^{2t}[/tex] is
[tex]e^{3t}*e^{2t}=\displaystyle\int_0^t e^{3(t-\tau)}e^{2\tau}\,\mathrm d\tau[/tex]
[tex]=\displaystyle e^{3t} \int_0^t e^{-\tau}\,\mathrm d\tau[/tex]
[tex]=e^{3t}(1-e^{-t})=e^{3t}-e^{2t}[/tex]
We end up with
[tex]\mathscr L^{-1}\left\{\dfrac4{(s-3)(s-2)}\right\}=\boxed{4(e^{3t}-e^{2t})}[/tex]
