partial fraction madness, continuation

- satellite73

partial fraction madness, continuation

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- anonymous

method for finding \(b\) in \[\frac{5x^2+20x+6}{x(x+1)^2}=\frac{a}{x}+\frac{b}{x+1}+\frac{c}{(x+1)^2}\]

- mathslover

in terms of a and c and x ?

- anonymous

@mahmit2012 method with derivatives

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## More answers

- anonymous

what's wrong with ordinary method of finding partial differential?

- anonymous

@mahmit2012 had a snap method, but i don't understand it

- mathslover

just a question:
do wolfram has solution?

- anonymous

they have something ugly...

- anonymous

we get \(a=\frac{6}{1}=6\), and \(c=\frac{-9}{-1}=9\) but
\[b=5+0-\frac{6}{(-1)^2}=-1\] step i don't get

- anonymous

http://www.wolframalpha.com/input/?i=integrate+%285x%5E2%2B20x%2B6%29+%2F+%28x%5E3%2B2x%5E2%2Bx%29

- anonymous

oh the original question was an integral, btw.

- anonymous

something to do with a derivative but i don't see it

- anonymous

what's your method for B

- anonymous

is there link to where it was done in snappier way?

- anonymous

looks like the method was, take \[5x^2+20x+6\] divide by \(x\) get
\[5x+20+\frac{x}{6}\] take the derivative get
\[5-\frac{6}{x^2}\] and then evaluate at \(x=-1\)
why this works is going to bug me

- anonymous

but is sure is snapp!

- anonymous

I think we can solve this making three equations....

- anonymous

i meant did you have your own way, satellite?

- anonymous

a slower... sane approach

- anonymous

i plugged in A and C.. expanded it all out... plugged in 1 for x and got B=-3

- anonymous

ooooooooooooooh!!!!
multiply both sides by \((x+1)^2\)!!!!!

- anonymous

but that's not right i guess

- anonymous

wowee zowee

- anonymous

how cool is that? i love it

- anonymous

@alienbrain yes i wrote my method before
solved \(6+B=5\) but it required some visualization

- anonymous

|dw:1344137128472:dw|

- anonymous

it is already solved, i was trying to understand @mahmit2012 approach

- anonymous

now i get it, maybe i can remember is
@alienbrain do not multiply out, you will make an algebra mistake, it is a pain
we know \(6+B=5\) and so \(B=-1\)

- anonymous

|dw:1344137486006:dw|

- anonymous

@mahmit2012 my less sophisticated method was to visualize the coefficient of \(x^2\) as \(6+B\)

- anonymous

@satellite73 are u sure your answer is correct.... because I got
a=15
b=-10
and c=1

- anonymous

yeah i am sure

- anonymous

|dw:1344137568037:dw|

- anonymous

yes i see thanks
i didn't understand that you multiplied on both sides

- anonymous

|dw:1344137675922:dw|

- anonymous

i think i would not want to use this method if i had to take the derivative repeatedly

- anonymous

this is the shortest method to get cofficiant in partial fraction problem >

- anonymous

back, open study lagged me out

- anonymous

yeah it was getting laggy
hope we have worn this topic out sufficiently
we have the answer and a plethora of methods

- anonymous

so 5=6 + B are you just combining like terms there

- anonymous

good thing you can't throw a math book at me:)

- anonymous

ok lets go back back back

- anonymous

we know \(A=6\)

- anonymous

first term on the left is \(A(x+1)^2=6(x+1)^2\) and we also have a term that looks like \(Bx(x+1)\)

- anonymous

whatever all the other mess is when you multiply out, the only way to get \(x^2\) is from the first term, which will give you \(6x^2\) and from the second term which will give you \(Bx^2\)

- anonymous

on the other hand we know we have to end up with \(5x^2\)
that tells you \(6+B=5\) and so \(B=-1\)

- anonymous

ok, i got it... but the method seems to vary for each letter

- anonymous

i was hoping to avoid critical thinking and just be able to do it by a process :)

- anonymous

ok you can use one method if you like
you can multiply this mess out and solve a 3 by 3 system, but i try to avoid that if at all possible because it is a ton of work
try to look for values of \(x\) that will make everything zero
obviously that will not work all the time when you have repeated factors but i find it much easier to use that method as far as i can
once you have some of the constants, the others will be easier to find, even if you have a system of equations

- anonymous

if you have no repeated factors, the it always works

- anonymous

well, if they are linear

- anonymous

it is also very easy to make an algebra mistake when multiplying out and combining like terms, much few mistakes plugging in numbers and evaluating
in any case if you were going to use "one method" you would have ended up with the equation \(A+B=5\) but think of how much easier if you know \(A=6\) already
and we got \(A=6\) with almost no work

- anonymous

ya

- anonymous

ok enough partial fractions for one night
i wish @eliassaab was here to show another method
it has to do with taking the limit as \(x\to \infty\) which for a rational function you can pretty much eyeball, and then determining what a constant is from that

- anonymous

eh.. integral of 6/x ?

- mathslover

@vishweshshrimali5

- anonymous

oh nm

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