## El_Arrow one year ago i need help with partial fractions problem

1. El_Arrow

$\frac{ 2x }{ (x-1)^3 }$

2. El_Arrow

@zepdrix @ganeshie8

3. zepdrix

You have a repeated linear factor in the denominator: $$\large\rm (x-1)$$ is linear. So do you remember how to break that down? I'll give an example:$\large\rm \frac{4x+2}{(x+1)^2}=\frac{A}{x+1}+\frac{B}{(x+1)^2}$And then you would solve for A and B.

4. zepdrix

$\large\rm \frac{2x}{(x-1)^3}=\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{(x-1)^3}$Understand the initial setup? :O

5. El_Arrow

yeah

6. El_Arrow

i used x=1 and got 2 for c is it correct?

7. zepdrix

mmm yah that sounds right! :)

8. El_Arrow

getting the other 2 looks complicated

9. zepdrix

Hmm I think I see a neat trick for the other ones :) Obviously we'd like to avoid expanding everything right?

10. El_Arrow

yeah

11. El_Arrow

okay let me try

12. zepdrix

Oh I made a boo boo.. my bad :( $$\large\rm A\ne B$$ I forgot about the C=2 part on the end, woooops.. back up :O

13. zepdrix

Back to the drawing board >.< woops woops

14. zepdrix

I still like the idea of plugging in x=0, and x=-1 it will give you a system of 2 equations with A and B.

15. zepdrix

It's probably about the same amount of work as expanding though

16. El_Arrow

well i have the final answer at the back of the book and its |dw:1442723235969:dw|

17. El_Arrow

so c is not equal to 2

18. zepdrix

Really? Hmm wolfram is saying that we should have c=2 and b=2 https://www.wolframalpha.com/input/?i=partial+fraction+decomposition+%282x%29%2F%28x-1%29%5E3

19. zepdrix

maybe typo in the book :O

20. El_Arrow

i guess there is a typo in the book

21. El_Arrow

so how does it get B=2?

22. zepdrix

I keep running into mistakes when I try to plug values in. I'm just gonna expand it out and see if we can get the right values...

23. El_Arrow

okay im gonna try that too

24. zepdrix

$\large\rm 2x=A(x-1)^2+B(x-1)+C$$\large\rm 2x=A(x-1)^2+B(x-1)+2$$\large\rm 2x=A(x^2-2x+1)+B(x-1)+2$$\large\rm 2x=Ax^2-2Ax+A+Bx-B+2$Then, matching up the squares,$\large\rm 0x^2=Ax^2\qquad\implies\qquad 0=A$Matching up the first degree terms,$\large\rm 2x=-2Ax+Bx\qquad\implies\qquad 2=-2A+B$And matching up the constant terms,$\large\rm 0=A-B+2$

25. El_Arrow

so |dw:1442723741155:dw|

26. El_Arrow

thanks for helping me figuring this one out

27. zepdrix

Ya I guess we shoulda just expanded from the beginning D: wasn't as tough as i thought it would be