I can do these types of problems without any trouble. What I don’t get is the why behind it. Every single source I have seen says something along the lines of “this is what you do” but does not explain why. Any help?
Here is an example partially worked out.
x/[(x+2)^2 * (x-1) = A/(x+2) + B/(x+2)^2 + C/(x-1)
Where I can’t figure out is why do you have to use both x+2 and (x+2)^2 in the factored form?
Any help would be most appreciated.
You need to have three equations in three unknowns in order to solve the system uniquely. Three equations in two unknowns may not have a solution.
Interesting question. After thinking it over for a minute, I realized it comes down to the basic structure theory for modules over a PID. The PID in question is R (R is the reals, but any field will do) and the module is R mod the ideal generated by (x-1) and (x+2)^2 in your example. This module is isomorphic to a direct sum of cyclic modules, generated by R/(x-1), R/(x+2) and R/(x+2)^2 and you are finding the coordinates of x itself in that direct sum.
If you don’t know enough to understand that answer, I apologize for that, but that is the only answer you can give other than: it works.
Think of it the other way around. If you started with something like what you have on the right side of the equation, and you were looking for the lowest common denominator so that you could add them together, it’d be the denominator on the left side.
So, when you start with the fractional expression on the left, and you’re thinking “What sort of simpler fractions could add up to that?” the answer is what’s on the right.
The Wikipedia article on Partial Fractions looks like it might help answer this, although if you don’t understand abstract algebra you might not understand everything.
ignore this post
There are really two questions:
a) Why do I need all three of those terms – 1/(x+2), 1/(x+2)^2, and 1/(x-1) – instead of just two – (x+2) and (x-1)?
b) Why do I use those three terms in particular?
For a), the answer is relatively straightforward as I pointed out: with less than three terms you likely won’t have a solution, because you will have two equations in three unknowns. Of course, you could still use different terms and get a unique solution, e.g. 1/(x+2), 1/[(x+2)*(x-1)] and (x-1).
For b), you need to get into the idea of irreducibility and principle ideal domains and what have you, as you noted.
Actually, you could use only two fractions, but the fraction with the second degree denominator should have a first degree polynomial in the numerator:
(Ax+B)/(x+2)^2 + C/(x-1).