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

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El_Arrow
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ 2x }{ (x1)^3 }\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.5You have a repeated linear factor in the denominator: \(\large\rm (x1)\) 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.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.5\[\large\rm \frac{2x}{(x1)^3}=\frac{A}{x1}+\frac{B}{(x1)^2}+\frac{C}{(x1)^3}\]Understand the initial setup? :O

El_Arrow
 one year ago
Best ResponseYou've already chosen the best response.0i used x=1 and got 2 for c is it correct?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.5mmm yah that sounds right! :)

El_Arrow
 one year ago
Best ResponseYou've already chosen the best response.0getting the other 2 looks complicated

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.5Hmm I think I see a neat trick for the other ones :) Obviously we'd like to avoid expanding everything right?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.5Oh 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
 one year ago
Best ResponseYou've already chosen the best response.5Back to the drawing board >.< woops woops

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.5I 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
 one year ago
Best ResponseYou've already chosen the best response.5It's probably about the same amount of work as expanding though

El_Arrow
 one year ago
Best ResponseYou've already chosen the best response.0well i have the final answer at the back of the book and its dw:1442723235969:dw

El_Arrow
 one year ago
Best ResponseYou've already chosen the best response.0so c is not equal to 2

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.5Really? 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%28x1%29%5E3

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.5maybe typo in the book :O

El_Arrow
 one year ago
Best ResponseYou've already chosen the best response.0i guess there is a typo in the book

El_Arrow
 one year ago
Best ResponseYou've already chosen the best response.0so how does it get B=2?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.5I 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
 one year ago
Best ResponseYou've already chosen the best response.0okay im gonna try that too

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.5\[\large\rm 2x=A(x1)^2+B(x1)+C\]\[\large\rm 2x=A(x1)^2+B(x1)+2\]\[\large\rm 2x=A(x^22x+1)+B(x1)+2\]\[\large\rm 2x=Ax^22Ax+A+BxB+2\]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,\[\large\rm 0=AB+2\]

El_Arrow
 one year ago
Best ResponseYou've already chosen the best response.0so dw:1442723741155:dw

El_Arrow
 one year ago
Best ResponseYou've already chosen the best response.0thanks for helping me figuring this one out

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.5Ya I guess we shoulda just expanded from the beginning D: wasn't as tough as i thought it would be
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