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.
 A: Starting from here $$\frac{9}{(1+2x)(2-x)^2} =\frac{A(2-x)^2 +B(1+2x)(2-x)+C(1+2x)}{(1+2x)(2-x)^2}$$
ignore the bottom lines:
$$9=A(2-x)^2+B(1+2x)(2-x)+C(1+2x)$$
Substitute values of $x$ to make brackets zero:
$$x=2\implies9=5C$$
$$x=-\frac12\implies 9=A(2+\frac12)^2$$
Compare coefficients:
$$x^2\implies0=A-2B\implies B=...$$
Putting $x=0$ gives a simple equation also.
This is more efficient than setting up simultaneous equations.
A: 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 ?
A: The other two answers have given you the standard way to solve this, but I have always found it easier to build up complicated fractions step-by-step.
In this case it is very easy to see that
$$
\frac{1}{(1+2x)(2-x)}=\frac{1}{5} \left[ \frac{2}{1+2x}+\frac{1}{2-x} \right]. \tag{*}
$$
So multiply this by $\frac{1}{2-x}$ and get
$$
\frac{1}{(1+2x)(2-x)^2}=\frac{1}{5} \left[ \frac{2}{(1+2x)(2-x)}+\frac{1}{(2-x)^2} \right].  
$$
Now all we need to do is use $(*)$ to deal with the first term on the right-hand side and we are done.
A: You say
"I understand that it will turn into-
$\frac{9}{(1+2x)(2−x)^2}= \frac{A}{1+2x}+\frac{B}{(1−x)}+\frac{C}{1−x^2}$".
But this is just wrong!  For one thing, the "2- x" has mysteriously turned into "1- x".  For another you have "$1- x^2$" where you should have "$(2- x)^2$".
You need $\frac{9}{(1+ 2x)(2- x)^2}= \frac{A}{1+ 2x}+ \frac{B}{2- x}+ \frac{C}{(2- x)^2}$.
We have three unknown values so need three equations to solve for A, B, and C.  There are many ways to get them.  I prefer getting rid of the fractions by multiplying by $(1+ 2x)(1- x)^2$:
$9= A(2- x)^2+ B(1+ 2x)(2- x)+ C(1+ 2x)$
Now we can get three equations by taking x to be three different numbers.  Choosing x= 2 and x= -1/2 make the equation very easy:
If x= 2, 2- x= 0 and we have 9= C.
If x= -1/2, 1+ 2x and we have 9= 9A/4 so A= 4.
0 is also easy- if x= 0 we have 9= 4A+ 2B+ C= 36+ 2B+ 9.
2B= -36 so B= -18.
$\frac{9}{(1+ 2x)(2- x)^2}= \frac{4}{1+ 2x}- \frac{18}{2- x}+ \frac{9}{(2- x)^2}$
