Concept Decomposing of Partial Fractions

[babby]

I have a few questions about decomposing partial fractions. I know how to solve these problems, but I just don't understand why I'm doing some of the things.

1. Why does equating coefficients work? I don't understand the idea behind it.

2. When you are decomposing fractions into constants

EX:

1/(x-1)(x-2)^2 = A/(x-1) + B/(x-2) + C/(x-2)^2

Why do you have to repeat (x-2), instead of just putting B/(x-2)^2?

Thanks.

 

[g_edgar]

You could, instead, do this:
[tex]\frac{A}{x-1} + \frac{Bx+C}{(x-2)^2}[/tex].
The point is that the numerator can be anything with degree less than the degree of the denominator.

 

[babby]

You could, instead, do this:
[tex]\frac{A}{x-1} + \frac{Bx+C}{(x-2)^2}[/tex].
The point is that the numerator can be anything with degree less than the degree of the denominator.

Ah, that was what I was originally thinking, but forgot the variable. But is there a reason why you have to repeat the function when doing it the other way? I don't understand how it works. Is it because the first part of the function:

B/(x-2) eliminates the need for a numerator with one degree less than the denominator seen in

C/(x-2)^2?

 

[babby]

OK, I understand that part now, but I have another question:

Can somebody explain to me why equating coefficients work?

Example:
8x^3+13x = Ax^3 + 2Ax + Bx^2 + 2B + Cx + D

expanded into:

8x^3 + 13x = Ax^3 + Bx^2 + (2A+C)x + 2(B+D)

where A,B,C,D are constants.

Why does 8 = A; 0 = B; 13 = 2A + C; etc.

I know they have same power variables, but why does this actually work? Thanks!

 

[MisterX]

The technique involves the assumption that for the correct values of unknowns such as A, B, and C, both sides of the equation are equivalent for all x. This assmption would still hold after manipulation to put both sides in polynomial form. It should be obvious that two polynomial functions of x are equal for all x if the corresponding coefficients are equal.