You don't need lineair and quadratic expressions in the nominators.
When the denominator is either lineair (ax+b), or lineair to a certain power (ax+b)^n, the the proposed nominator is just a number.
It's only when the denominator is a quadratic with discriminant < 0 that the nominators is proposed to be lineair. Quadratics in the nominator never appear.
I have the nominators, they aren't a problem. I have run this several times and have come up with different answers each time.
When I said negative, I meant to say minus.
You have posted a number of times that you get "negative answers" (which isn't necessarily wrong), that you get two different answers, and that you are lost but you still haven't shown what you have done so we could point out possible mistakes!
You want to find A, B, C, D, E so that
[tex]\frac{-8x^3- 2}{x^2(x-1)^3}= \frac{A}{x}+ \frac{B}{x^2}+ \frac{C}{x-1}+ \frac{D}{(x-1)^2}+ \frac{E}{(x-1)^3}[/tex]
TD wrote the left side of that as a single fraction but I think it is easiest to multiply both sides of that by x^{2}(x-1)^{3} to get
[tex]-8x^3- 2[/tex]= Ax(x-1)^3+ B(x-1)^3+ Cx^2(x-1)^2+ Dx^2(x-1)+ Ex^2[/itex]
The easy thing to do is take x= 0 so that
[tex]-2= -B[/tex] or B= 2
and then take x= 1 so that
[tex]-10= E[/tex].
Since those are the only two x values that make things on the right side 0, you now have two different ways to find A, C, D.
(a) Multiply the whole thing out and combine like powers of x
(b) Use simple values for x, like -1, 2, -2, to get three equation for A, C, D.
Related Threads for: Partial fraction decomposition