Equation is:
L{${(t+1)}^2e^t$}
By using the properties of Laplace, I got this result $$\frac{-2}{(s-1)^3} +\frac{2}{(s-1)^2} + \frac{1}{(s-1)}$$ How do I plot the zeros and poles? Also, what does it mean to plot on s plane, How is it different from the normal plane.
$s$ plane is nothing but complex plane where you plot poles and zeros. Zeros are the values where $L(s)=0$ and Poles are the values where $L(s)\mapsto \infty$
Upon simplifying $$\frac{-2}{(s-1)^3} +\frac{2}{(s-1)^2} + \frac{1}{(s-1)}=\frac{s^2-3}{\left(s-1\right)^3}$$
So the pole is $-1$ And zeros are $\pm \sqrt{3}$
While plotting you denote poles with $\times$ and zeros with $O$ So there will be a $x$ on imaginary axis at $-1$ and two $O$ at $\pm \sqrt{3}$ on real axis.