Partial fraction decomposition

[eric_999]

Alright, I am here again with another question...

When I have a rational function, let's say (x+4)/(x-2)(x-3) I rewrite it like A/(x-2) + B(x-3) and then solve it for A & B. But when we have for e.g (x^2 + 3x + 2)/(x(x^2 +1 )) the book tells me to rewrite it like:
A/x + (Bx + C)/(x^2 + 1), and then solve for A & B. I understand that the term x^2 +1 cannot be further decomposed (at least not if we only consider real numbers). However feel I don't get everything.

For example if I instead try to rewrite it on the form A/x + B/(x^2 + 1), so A(x^2 + 1) + Bx = x^2 +3x + 2, so A = 1, B = 3, and A = 2, which is of course impossible. On the other hand I can see that the other form described above (which the book tells me to use) works fine.

The problem is that with different rational functions I might be able to try different strategies and just see which one works out, but I feel i don't understand it the way I want to. In the book they simply say "the rational function P(x)/Q(x) can be expressed as a sum of partial fractions like this: ... but we don't explain it further because this is not a course in algebra" I feel I need to really understand, not just memorize the techniques!

Thanks for help!

 

[HallsofIvy]

There are, in fact, a number of ways of finding the coefficients for the fractions. If you have
$$\frac{x+ 4}{(x- 2)(x- 3)}= \frac{A}{x- 2}+ \frac{B}{x- 3}$$

1. Do the addition on the right: multiply both numerator and denominator of the first fraction by x- 3 and the numerator and denominator of the second fraction by x- 2:
$$\frac{x+ 4}{(x- 2)(x- 3)}= \frac{A(x- 3)}{(x- 2)(x- 3)}+ \frac{B(x- 2)}{(x- 2)(x- 3)}$$
$$= \frac{Ax- 3A+ Bx- 2B}{(x- 2)(x- 3)}= \frac{(A+ B)x- (3A- 2B)}{(x- 2)(x- 3)}$$
so we must have A+ B= 1 and -3A- 2B= 4.

2. Multiply both sides by (x- 3)(x- 2):
$$x+ 4= A(x- 3)+ B(x- 2)= (A+ B)x- (3A+ 2B)$$
so that we have the same two equations.

3. After getting
$$x+ 4= A(x- 3)+ B(x- 2)$$
choose any two values you like for x to get two linear equations for A and B.

4. In particular, choosing x= 2 and x= 3 gives very simple, separated, equations: 6= -A and 7= B.

 

[verty]

Here is a pdf that makes it nice and clear.

The proof of these decompositions is not going to be too interesting, it'll just show that in each case, for the right coefficients, the simplest form of the numerator is the one given by the theorem.

 

[eric_999]

Thanks!