Confusion for two-sided signal Given a continuous LTI system with transfer function
$$H(s)= -\frac{2s}{(s+6)(s+2)}$$

Plot the location of the pole(s) and zero(s)
Find all possible regions of convergence
From the problem above find the impulse response

Here's my attempt

The poles are $s_1=-6$ and $s_2=-2$ and the zero is $s_3=0$

The all possible roc-s are

Using inverse laplace transform I found $h(t)=e^{-2t}-3e^{-6t}$. For the right-sided signal I multiply $h(t)$ with $u(t)$ and I got $h_{RS}(t)=(e^{-2t}-3e^{-6t})u(t)$ and for the left-sided signal I multiply $h(t)$ with $u(-t)$ and I got $h_{LS}(t)=(e^{-2t}-3e^{-6t})u(-t)$.

My confusion is at the two-sided signal. My solution is $h_{TS}(t)=e^{-2t}-3e^{-6t}$ and my friend's solution is $h_{TS}(t)=-e^{-2t}u(-t) - 3e^{-6t}u(t)$. When I plotted the signal both are a two-sided signal. Since my professor didn't submit the solution we didn't know who's right or wrong.
Thank you.
Given LTI system with transfer function :
$$H(s) = \frac{ -2 s}{(s+6)(s+2) } \tag{1}$$
which has the poles $s = \{ -2 , -6 \} $, and zeros $ s = 0 $.
Apply partial fraction expansion :
$$H(s) = \frac{1}{s+2} - \frac{3}{s+6} \tag{2}$$
Based on pole locations, there will be three ROCs,

ROC1 :  -$2 &lt; \mathcal{Re}\{s\} &lt; \infty $  ; stable
ROC2 : $-\infty &lt; \mathcal{Re}\{s\} &lt; -6$  ; unstable
ROC3 : $ -6 &lt; \mathcal{Re}\{s\} &lt; -2$   ; unstable

The impulse responses found from (2) and the chosen ROC , with the fact that
$$\mathcal{L}\{ e^{at}u(t) \} \longleftrightarrow \frac{1}{s-a} ~~,~~ \mathcal{Re}\{s\} &gt;\mathcal{Re}\{a\} \tag{3}$$
$$\mathcal{L}\{ -e^{at}u(-t) \} \longleftrightarrow \frac{1}{s-a} ~~,~~ \mathcal{Re}\{s\} &lt;\mathcal{Re}\{a\} \tag{4}\\\\$$

ROC1 :  -$2 &lt; \mathcal{Re}\{s\} &lt; \infty $  ;  right-sided  : $h(t) = (e^{-2t}-3 e^{-6t}) u(t)\\\\$
ROC2 : $-\infty &lt; \mathcal{Re}\{s\} &lt; -6$  ; left-sided : $h(t) = (-e^{-2t}+3 e^{-6t}) u(-t)\\\\$
ROC3 : $ -6 &lt; \mathcal{Re}\{s\} &lt; -2$   ; two-sided : $h(t) = -e^{-2t}u(-t)-3 e^{-6t}u(t)$.