# Partial fractions, proper fraction with a perfect square factor

I think I have figured why the following is the case, I need however a confirmation that it is indeed so.

Consider:

$\displaystyle{\frac{x-1}{(x+1)(x-2)^2} \equiv \frac{A}{x+1} + \frac{Bx+C}{(x-2)^2}} \ \ (i)$

Now the book is showing the following step: let $C=-2B+D$, so that

$\displaystyle{\frac{Bx+C}{(x-2)^2} \equiv \frac{Bx-2B+D}{(x-2)^2} \equiv \frac{B}{x-2}+\frac{D}{(x-2)^2}} \ \ (ii)$

I was thinking about the peculiar choice of $C$ as $-2B + D$, i.e. the relationship of constant $C$ has been expressed of already present constant $B$ (they did not express $C$ as some other arbitrary constants, but rather a combination of existing constant $B$ and some arbitrary constant $D$). But the choice leads to the simplest proper partial fraction form of the initial fraction, and that is why $C = -2B + D$. Is this correct?

It also means that if you express the initial fraction in the RHS form of (i), then you can easily express it in the simplest partial fraction form by keeping in mind that $C=-2B+D$ ?

Are there any cases where this would not work?

• This trick is made in order that $Bx+C=B(x-2)+(C+2B)$ simplifies partly with the denominator which is $(x-2)^2$. In fact, this is the kind of trick you will often use. Is this better ? If not, just post. – Claude Leibovici Jul 19 '15 at 8:36
• Ohhhh. Nice :) thank you! – i squared - Keep it Real Jul 19 '15 at 8:51
• You are very welcome ! – Claude Leibovici Jul 19 '15 at 8:53

## 1 Answer

The choice of $C=-2B+D$ is for this particular example, and is used to illustrate the fact that with repeated factors, you express the fraction as separate terms with a single letter (no $x$ terms) in the numerator, and successive powers of the factor in the denominator.

So, for example, you would decompose $$\frac{1}{(x-a)^3}$$ as $$\frac{A}{(x-a)}+\frac{B}{(x-a)^2}+\frac{C}{(x-a)^3}$$ and so on.