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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.

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    $\begingroup$ you need to edit your question - the denominators are supposed to have $(2-x)$ terms not $(1-x)$. $\endgroup$
    – David Quinn
    Commented Apr 5, 2022 at 11:01
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    $\begingroup$ You also have written $1-x^2$ when it should have been $(1-x)^2$ [Except as @DavidQuinn says they should be powers of $2-x$]. $\endgroup$
    – ancient mathematician
    Commented Apr 5, 2022 at 11:24
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    $\begingroup$ You have changed the question after receiving two perfectly good answers. That doesn't seem very fair, does it? $\endgroup$
    – ancient mathematician
    Commented Apr 6, 2022 at 9:12
  • $\begingroup$ Sorry, I've figured out the answers from these examples, and they helped me receive an answer, it was careless of me to post a questions without checking first. The original question was different. $\endgroup$
    – verygood101
    Commented Apr 8, 2022 at 11:53

4 Answers 4

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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.

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  • $\begingroup$ I see, but how would we make the bracket 0 for $(1-x)^2$ ? Would I just expand or is there another method to this? $\endgroup$
    – verygood101
    Commented Apr 6, 2022 at 8:13
  • $\begingroup$ In your example there are only 2 values which make brackets zero. When you've run out of values, either compare coefficients or just select another $x$ value such as 0 $\endgroup$
    – David Quinn
    Commented Apr 6, 2022 at 8:55
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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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  • $\begingroup$ How would you get those sets of equations? $\endgroup$
    – verygood101
    Commented Apr 6, 2022 at 7:19
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    $\begingroup$ You develop the numerator of the RHS and then you compare the coefficients of $x^0,x$ and $x^2$. On the LHD, the coefficient of $x^0$ is $9$, the coefficient of $x$ is $0$ and the coefficient of $x^2$ is also $0$. $\endgroup$
    – SacAndSac
    Commented Apr 6, 2022 at 7:27
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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.

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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.

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

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