This question is specifically about zeroes in $X_1(s)$ and $X_3(s)$ in the shown diagrams. Attached snapshot contains complete problem & solution.
Since the poles($\color{blue}{\times}$) are at $-1,1$, easy to guess that the denominator of the laplace transform will be $(s+1)(s-1)$. I think this is the reason they're writing the partial fraction form as $$\dfrac{A}{s+1}+\dfrac{B}{s-1}$$
I get so far. No issues.
But next they're setting $A=B$ for a single zero($\color{blue}{\circ}$) at $s=0$ condition, why are they doing this ? Numerator can have function of $s$ as a factor, right ?
Similarly they're setting $A=-B$ for no zeroes condition. How does this work ?
(To find zeroes, we set the numerator to 0 and solve. I'm wondering how setting A=B is equivalent to setting the numerator to 0.)
$$X_3(s) = \dfrac{A}{s+1}+\dfrac{A}{s-1}= \dfrac{2As}{(s+1)(s-1)}$$
So $X_3(0) = 0$.
$$X_1(s) = \dfrac{A}{s+1}+\dfrac{-A}{s-1}= \dfrac{-2A}{(s+1)(s-1)}$$
So $X_1(s)$ only has zeros at $\infty$.