I have the following expression to get its laplace transfer: $$e^{2t}(3t-3t^2)$$
Is it ok to just calculate the transfer of each term then multiply the result? I calculated the expression above like this but it is different than the answer in my book:
$${\frac{1}{s-2}}*\frac{3}{s^2}-\frac{6}{s^3}$$ $$\frac{1}{s-2}*\frac{3s^3-6s^2}{s^5}$$ $$\frac{3s^3-6s^2}{(s-2)s^5}$$
We have
$$e^{2t}(3t-3t^2)=3e^{2t}(t-t^2)$$
$$\mathcal{L}(t-t^2)=\mathcal{L}(t)-\mathcal{L}(t^2)=\frac{1}{s^2}-\frac{2}{s^3}$$
You cannot simply multiply the transforms as you wanted (try it on $t^2=t *t$ as a counter example). However, using David Mitra's shift rule, we have
$$3\mathcal{L}(e^{2t}(t-t^2))=3\left(\frac{1}{(s-2)^2}-\frac{2}{(s-2)^3}\right)=\frac{3(s-4)}{(s-2)^3}$$