I'm not a close friend of partial fractions but I'll do this as follows:
Multiply this last result by and you'll get the desired partial fraction.
(Of course someone else will be give you a solution which involves your procedure.)
Would very much appreciate verification of solution to partial fraction decomposition question.
Find the partial fraction decomposition of x^4/(x-1)^3
Since degree of numerator is greater than denominator ... long division gives
3x^3-3x^2+x. Since degrees are same, another long division gives:
6x^2 -8x +3.
With A/(x-1) + B/(x-1)^2 + C/(x-1)^3
My solution is: A=6, B=4 and C=1
But my "check" does not compute.
Appreciate any help or redirection.
THAT WAS PURE AWESOME
Hmmm... now I need to say something relevent to stop this post being spam.
I think that the problem is in your checking because my check worked fine:But my "check" does not compute.
A(x-1)^2+B(x-1)+C = 6x^2-8x+3
6(x^2-2x+1)+4(x-1)+1 = 6x^2-8x+3
6x^2-12x+6+4x-4+1 = 6x^2-8x+3
6x^2-8x+3 = 6x^2-8x+3
Your method all looks sound.