<s>[INST] Finding the partial fractions decomposition of $\frac{9}{(1+2x)(2-x)^2} $ So this is basically my textbook work for my class, where we are practicing algebra with partial fractions.
I understand the basics of decomposition, but I do not understand how to do it when then the denominator is a power of $x^2$?
e.g. this question - $$\frac{9}{(1+2x)(1-x)^2} $$
I understand that it will turn into- $$\frac{9}{(1+2x)(2-x)^2} = \frac {A}{1+2x} + \frac {B}{(1-x)}+ \frac {C}{(1-x)^2}$$ and then it will become
$$\frac{9}{(1+2x)(1-x)^2} =\frac{A(1-x)^2 +B(1+2x)(1-x)+C(1+2x)}{(1+2x)(1-x)^2}$$
but what do you do once you are at this step? The example on the textbook isn't very clear, so if anyone could tell me what I do after doing this, and why that is the case, I would be very thankful.
 [/INST] You have $$\frac{9}{(1+2x)(2-x)^2} =\frac{A(2-x)^2 +B(1+2x)(2-x)+C(1+2x)}{(1+2x)(2-x)^2}$$
By equaling coefficients of the terms $x^0$, $x^1$ and $x^2$, you have the following equations to solve :
$$
\begin{cases}
4A+2B+C=9  \\
-4A+3B+2C=0\\
A-2B=0
\end{cases}$$
Can you continue from here ?
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