## satellite73 Group Title partial fraction madness, continuation 2 years ago 2 years ago

1. satellite73 Group Title

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 Group Title

in terms of a and c and x ?

3. satellite73 Group Title

@mahmit2012 method with derivatives

4. Libniz Group Title

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

5. satellite73 Group Title

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

6. mathslover Group Title

just a question: do wolfram has solution?

7. alienbrain Group Title

they have something ugly...

8. satellite73 Group Title

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. alienbrain Group Title
10. alienbrain Group Title

oh the original question was an integral, btw.

11. satellite73 Group Title

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

12. alienbrain Group Title

13. Libniz Group Title

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

14. satellite73 Group Title

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. satellite73 Group Title

but is sure is snapp!

16. sauravshakya Group Title

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

17. alienbrain Group Title

i meant did you have your own way, satellite?

18. alienbrain Group Title

a slower... sane approach

19. alienbrain Group Title

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

20. satellite73 Group Title

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

21. alienbrain Group Title

but that's not right i guess

22. satellite73 Group Title

wowee zowee

23. satellite73 Group Title

how cool is that? i love it

24. satellite73 Group Title

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

25. sauravshakya Group Title

|dw:1344137128472:dw|

26. satellite73 Group Title

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

27. satellite73 Group Title

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. mahmit2012 Group Title

|dw:1344137486006:dw|

29. satellite73 Group Title

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

30. sauravshakya Group Title

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

31. satellite73 Group Title

yeah i am sure

32. mahmit2012 Group Title

|dw:1344137568037:dw|

33. satellite73 Group Title

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

34. mahmit2012 Group Title

|dw:1344137675922:dw|

35. satellite73 Group Title

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

36. mahmit2012 Group Title

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

37. alienbrain Group Title

back, open study lagged me out

38. satellite73 Group Title

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

39. alienbrain Group Title

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

40. alienbrain Group Title

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

41. satellite73 Group Title

ok lets go back back back

42. satellite73 Group Title

we know $$A=6$$

43. satellite73 Group Title

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. satellite73 Group Title

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. satellite73 Group Title

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. alienbrain Group Title

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

47. alienbrain Group Title

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

48. satellite73 Group Title

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. satellite73 Group Title

if you have no repeated factors, the it always works

50. satellite73 Group Title

well, if they are linear

51. satellite73 Group Title

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. alienbrain Group Title

ya

53. satellite73 Group Title

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. alienbrain Group Title

eh.. integral of 6/x ?

55. mathslover Group Title

@vishweshshrimali5

56. alienbrain Group Title

oh nm