## RolyPoly 3 years ago Is that correct? Resolve into partial fraction$\frac{x^2}{x^4-1}$$\frac{x^2}{x^4-1}=\frac{x^2}{(x^2+1)(x-1)(x+1)}$ $A(x-1)(x+1) + B(x^2+1)(x-1) + C(x^2+1)(x+1) = x^2$$(B+C) x^3+(A-B+C)x^2 +(B+C)x -A-B+C =x^2$So, B+C = 0, A-B+C=1, A+B-C = 0 And I got A = 1/2, B=-1/4, C=1/4 So, $\frac{x^2}{x^4-1} = \frac{1}{2(x^2+1)} - \frac{1}{4(x+1)} +\frac{1}{4(x-1)}$

1. zepdrix

|dw:1350710674124:dw|

2. .Sam.

@RolyPoly good to go.

3. RolyPoly

@zepdrix Yes! That should be the case, but I can't get the answer by solving that equation :(

4. zepdrix

Hmmm :c

5. RolyPoly

Continue from that step; (A+C+D)x^3 + (B+C-D)x^2 + (-A+C+D) + (-B+C-D) = x^2 So, A+C+D = 0 --- (1) (B+C-D) = 1 --- (2) -A+C+D =0 --- (3) -B+C-D = 0 --- (4) (1)-(3): A=0 (3)-(4): -A+B = 0 => B = 0 Here comes the problem!

6. zepdrix

|dw:1350711031547:dw| Hmmmm we have 4 equations and 4 unknowns. This should be solvable. It might be a bit of work though :o

7. zepdrix

Oh hmm

8. zepdrix

|dw:1350711417164:dw| Hmm maybe just try combining them a different way? <:O I came up with this, let me know if my math was incorrect.

9. zepdrix

I'm also getting A=0, I don't think that's a problem though. Since I was able to get a B term.

10. zepdrix

Oh i see the problem, (3)-(4) = (-A+C+D) - (-B+C-D) = -A + B +2D = 0

11. RolyPoly

Uh-huh! My mistake!! Thanks :)

12. zepdrix

Yah these things suck :) lol

13. RolyPoly

(2) - (4): B = 1/2 Yay!