<s>[INST] Why do you need two fractions for partial fraction decomposition with repeated factors? For example, suppose my denominator contains $(x - 1)(x - 1)$.
I know I need two fractions, one with $(x - 1)$ and one with $(x - 1)^2$ as the denominator. But I'm looking for a deeper reason as to why. It makes sense when you go through and get a common denominator that it will all work out in the end, but I just want a good explanation for it.
 [/INST] You don't actually need two fractions for the squared factor. If for example you wanted to express $\frac{4x}{(x+1)(x-1)^2}$ in partial fractions, you could express it as $$\frac{4x}{(x+1)(x-1)^2}\equiv\frac{a}{x+1}+\frac{bx+c}{(x-1)^2}$$ This gives you $$4x\equiv a(x-1)^2+(bx+c)(x+1)\equiv (a+b)x^2+(-2a+b+c)x+(1+c)$$and you can then solve the equations:
1)$\ \ a+b=0$
2)$\ -2a+b+c=4$ 
3)$\ c=0$
for $a,b$ and $c$ to get$$\frac{4x}{(x+1)(x-1)^2}\equiv \frac{-\frac{4}{3}}{x+1}+\frac{\frac{4}{3}x}{(x-1)^2}$$The point is that you have three unknowns ($a,b$and $c$) and three equations, which will generally give a solution.
Were you to assume$$\frac{4x}{(x+1)(x-1)^2}\equiv\frac{a}{x+1}+\frac{b}{(x-1)^2}$$in this case you would get$$4x\equiv a(x-1)^2+b(x+1)\equiv(a+b)x^2+(-2a+b)x+(1+b)$$
You then have to solve
1)$\ a+b=0$
2)$\ -2a+b=4$
3)$\  1+b=0$
This is three equations in two variables and has no solution.
Trying $$\frac{4x}{(x+1)(x-1)^2}\equiv\frac{a}{x+1}+\frac{b}{(x-1)^2}+c$$ would not help because then after clearing the denominator you finish up with a cubic, so you still have one more equation than you have variables. 
Assuming $$\frac{4x}{(x+1)(x-1)^2}\equiv\frac{a}{x+1}+\frac{b}{(x-1)}+\frac{c}{(x-1)^2}$$is usually the most convenient.
If you want to split a rational function into a sum of reciprocals of polynomials then this last form is necessary, but only possible if the denominator of the function you wish to split (in its lowest terms) is a product of linear factors. 
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