Finding the partial fractions decomposition of $\frac{9}{(1+2x)(2-x)^2} $

Question

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.

Answer 1

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 ?

Answer 2

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.

Answer 3

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.

$\frac{9}{(1+ 2x)(2- x)^2}= \frac{4}{1+ 2x}- \frac{18}{2- x}+ \frac{9}{(2- x)^2}$

Answer 4

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.