Question aboutPartial Fractions for example:
$$\frac{{{x^2} + 4}}{{x\left( {x + 2} \right)\left( {3x - 2} \right)}}\, = \frac{A}{x} + \frac{B}{{x + 2}} + \frac{C}{{3x - 2}}$$
first method is:
$${x^2} + 4 = A\left( {x + 2} \right)\left( {3x - 2} \right) + Bx\left( {3x - 2} \right) + Cx\left( {x + 2} \right)$$
but it is hard and take much time to find A,B,C
second method(Substituting the roots, or "zeros") is:
\begin{align*}x & = 0 \,\,\,\,\, : & \hspace{0.5in}4 & = A\left( 2 \right)\left( { - 2} \right) & \hspace{0.5in} & \Rightarrow  & \hspace{0.25in}A & =  - 1\\ x & =  - 2 : & \hspace{0.5in}8 & = B\left( { - 2} \right)\left( { - 8} \right) & \hspace{0.25in}&\Rightarrow & \hspace{0.25in}B & = \frac{1}{2}\\ x & = \frac{2}{3}\,\, : & \hspace{0.5in}\frac{{40}}{9} & = C\left( {\frac{2}{3}} \right)\left( {\frac{8}{3}} \right) & \hspace{0.25in} & \Rightarrow &  \hspace{0.25in}C & = \frac{{40}}{{16}} = \frac{5}{2}\end{align*}
it is better and easier method for Partial Fractions. My question is why this method works? can you prove this method?

Your "second method" is the next step of the first method. At each of the zeroes of $$x(x+2)(3x-2)$$ only one of the ABC terms is non-zero so you can solve for each ABC easily.
