Answer
The roots are $s=-2$ and $-4 .$ Thus
$$X(s)=\frac{1-e^{-3 s}}{(s+2)(s+4)}$$
Let
$$F(s)=\frac{1}{(s+2)(s+4)}=\frac{1}{2}\left(\frac{1}{s+2}-\frac{1}{s+4}\right)$$
so $$f(t)=\frac{1}{2}\left(e^{-2 t}-e^{-4 t}\right)$$
From Property 6 of the Laplace transform,
$$x(t)=\frac{1}{2}\left(e^{-2 t}-e^{-4 t}\right)-\frac{1}{2}\left[e^{-2(t-3)}-e^{-4(t-3)}\right] u_{s}(t-3)$$
Work Step by Step
The roots are $s=-2$ and $-4 .$ Thus
$$X(s)=\frac{1-e^{-3 s}}{(s+2)(s+4)}$$
Let
$$F(s)=\frac{1}{(s+2)(s+4)}=\frac{1}{2}\left(\frac{1}{s+2}-\frac{1}{s+4}\right)$$
so $$f(t)=\frac{1}{2}\left(e^{-2 t}-e^{-4 t}\right)$$
From Property 6 of the Laplace transform,
$$x(t)=\frac{1}{2}\left(e^{-2 t}-e^{-4 t}\right)-\frac{1}{2}\left[e^{-2(t-3)}-e^{-4(t-3)}\right] u_{s}(t-3)$$
