## satellite73 4 years ago partial fraction madness, continuation

1. anonymous

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

2. mathslover

in terms of a and c and x ?

3. anonymous

@mahmit2012 method with derivatives

4. anonymous

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

5. anonymous

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

6. mathslover

just a question: do wolfram has solution?

7. anonymous

they have something ugly...

8. 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

9. anonymous
10. anonymous

oh the original question was an integral, btw.

11. anonymous

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

12. anonymous

13. anonymous

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

14. 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

15. anonymous

but is sure is snapp!

16. anonymous

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

17. anonymous

i meant did you have your own way, satellite?

18. anonymous

a slower... sane approach

19. anonymous

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

20. anonymous

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

21. anonymous

but that's not right i guess

22. anonymous

wowee zowee

23. anonymous

how cool is that? i love it

24. anonymous

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

25. anonymous

|dw:1344137128472:dw|

26. anonymous

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

27. 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 $$B=-1$$

28. anonymous

|dw:1344137486006:dw|

29. anonymous

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

30. anonymous

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

31. anonymous

yeah i am sure

32. anonymous

|dw:1344137568037:dw|

33. anonymous

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

34. anonymous

|dw:1344137675922:dw|

35. anonymous

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

36. anonymous

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

37. anonymous

back, open study lagged me out

38. anonymous

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

39. anonymous

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

40. anonymous

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

41. anonymous

ok lets go back back back

42. anonymous

we know $$A=6$$

43. anonymous

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

44. 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 $$Bx^2$$

45. anonymous

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

46. anonymous

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

47. anonymous

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

48. 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

49. anonymous

if you have no repeated factors, the it always works

50. anonymous

well, if they are linear

51. 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

52. anonymous

ya

53. 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

54. anonymous

eh.. integral of 6/x ?

55. mathslover

@vishweshshrimali5

56. anonymous

oh nm