i need help with partial fractions problem

- El_Arrow

i need help with partial fractions problem

- schrodinger

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- El_Arrow

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

- El_Arrow

- 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.

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## More answers

- 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

- El_Arrow

yeah

- El_Arrow

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

- zepdrix

mmm yah that sounds right! :)

- El_Arrow

getting the other 2 looks complicated

- zepdrix

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

- El_Arrow

yeah

- El_Arrow

okay let me try

- 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

- zepdrix

Back to the drawing board >.< woops woops

- 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.

- zepdrix

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

- El_Arrow

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

- El_Arrow

so c is not equal to 2

- 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

- zepdrix

maybe typo in the book :O

- El_Arrow

i guess there is a typo in the book

- El_Arrow

so how does it get B=2?

- 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...

- El_Arrow

okay im gonna try that too

- 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,\[\large\rm 0=A-B+2\]

- El_Arrow

so |dw:1442723741155:dw|

- El_Arrow

thanks for helping me figuring this one out

- zepdrix

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

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