Algebra -- resolving into components I saw this in a math book. $$\frac{x}{(x-1)(2x+3)}=\frac{1}{(x-1)(2\cdot 1+3)}+\frac{-3/2}{(-3/2-1)(2x+3)}$$
I solve these things via $$\frac{x}{(x-1)(2x+3)}=\frac{A}{(x-1)}+\frac{B}{(2x+3)}$$
then solving for A and B..
I realise the above is a shortcut, but what I need to know is if I can use it everywhere or there are some pre-conditions (like, only valid for linear components ... meaning $x^2+1$ won't work here ... or something like that) for the above shortcut to work.
Please advise.
 A: The below is meant to be explanatory, not a formal proof of anything.
Have you thought about where the shortcut comes from? Take your method, and recombine the components:
$$\frac{A(2x+3)+B(x-1)}{(x-1)(2x+3)}=\frac{x}{(x-1)(2x+3)}\\
\implies A(2x+3)+B(x-1)=x$$
So now it's clear that setting x=1 and dividing through by (2x+3) gives $A$. This is exactly what the shortcut suggests.
Let's go more general:
$$\frac{f(x)}{g(x)h(x)}=\frac{i(x)}{g(x)}+\frac{j(x)}{h(x)}\\
\implies h(x)i(x)+g(x)j(x)=f(x)$$
So the shortcut approach of finding where $g(x)=0$ then dividing through by $h(x)$ does indeed give us the correct expression for $i(x)$.
Also, please note that we take that $\frac{f(x)}{g(x)h(x)}$ can be written as$\frac{i(x)}{g(x)}+\frac{j(x)}{h(x)}$ as an assumption.
Even more, there might be problems with dividing by $h(x)$. What if it is zero? These are real problems with my exposition. But I think it helps to explain that the rule can be more general than the example you provided, but is also subject to limitations.
