How to find A, B, C from equation mapping. Sir, Help me to find A,B and C when
$$\frac{A}x+\frac{B}{x^2}+\frac{C}{x-1}=\frac{x-2}{x^2(x-1)}$$
I try to do.
$(x(x-1)+(x-1)+x^2)/(x^2(x-1)) = (x-2)/(x^2(x-1))$
But I do not how to map A,B,C with coefficient. Any advise or guidance would be greatly appreciated, Thanks.
 A: This is an example of simplifying & solving the result of Partial fraction decomposition. As lab bhattacharjee's comment stated, you forgot to include the $A, B, C$ factors when you tried to solve the equation. Specifically, putting the left hand side (LHS) fractions each under a common denominator of $x^2(x-1)$, the numerator sum of the LHS must equal the numerator of the RHS. This gives by simplifying and putting items with the same power of $x$ together
$$\begin{equation}\begin{aligned}
A(x)(x-1) + B(x - 1) + C(x^2) & = x - 2 \\
Ax^2 - Ax + Bx - B + Cx^2 & = x - 2 \\
(A + C)x^2 + (B - A)x - B & = x - 2
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
The powers of $x$ terms must match on both sides for it to be true for all $x$, so you get
$$A+C = 0 \tag{2}\label{eq2A}$$
$$B - A = 1 \tag{3}\label{eq3A}$$
$$-B = -2 \implies B = 2 \tag{4}\label{eq4A}$$
This is a system of $3$ linear equations in $3$ unknowns. In this case, you can solve them quite easily by substituting \eqref{eq4A} into \eqref{eq3A} to get $A = 1$, then substituting this into \eqref{eq2A} to get $C = -1$.
A: Hint
As $$Ax(x-1)+B(x-1)+Cx^2=x-2$$ is an identity
set $x=0$ to find $-B=0-2$
Set $x=1$ to find $C=1-2$
Set $x=-1$
