## anonymous one year ago Partial Fraction decomposition: (x^2)/(x^4-2x^2-8)

1. anonymous

can you factor the denominator?

2. anonymous

Yes i factored denominator: (x^2+2)(x-2)(x+2)

3. anonymous

ok then we have $\frac{x^2}{(x^2+2)(x-2)(x+2)}$ $=\frac{Ax+B}{x^2+2}+\frac{C}{x-2}+\frac{D}{x-2}$ as a fist step

4. anonymous

actually i think in this case $$A=0$$ but it doesn't need to be

5. anonymous

would (A/x-2) + (B/x+2) + ((Cx+D)/(x^2+2)) be correct as well?

6. anonymous

now add $Ax+B(x+2)(x-2)+C(x^2+2)(x-2)+D(x^2+2)(x+2)=x^2$

7. anonymous

yes of course

8. anonymous

it doesn't matter since addition is commutative but since i spent all that time writing it in latex, would you mind using it so i don't have to start over?

9. anonymous

Yes that's completely fine. thank you

10. anonymous

so now there are a couple of ways to proceed the easiest way is probably to replace $$x$$ by $$2$$ and see what you get

11. anonymous

the first two terms drop out, you are left with $D(2^2+2)(2+2)=2^2$ and you can solve that for $$D$$

12. anonymous

that is really the basic gimmick is it clear what i did? you get $D=\frac{1}{6}$ pretty much in your head

13. anonymous

one minute i am reading your work

14. anonymous

ok take your time i am going to get a snack if it is not clear, ask

15. anonymous

I understand what you did. I worked the problem and got D=1/6

16. anonymous

ok good

17. anonymous

now we can use the same gimmick only replace $$x$$ yb $$-2$$

18. anonymous

Ok, so that cancels out the first and last term. I can now solve for C. C=-1/6?

19. anonymous

yes

20. anonymous

and now I plugin C and D into the first equation to solve for A & B?

21. anonymous

you can do that, but first lets think

22. anonymous

Ok..

23. anonymous

the first part is $(Ax+B)(x+2)(x-2)$

24. anonymous

if you multiply that out mentally you get an $$Ax^3$$ term there is not $$x^3$$ so $$A=0$$

25. anonymous

Sorry I do not understand why Ax^3 = 0

26. anonymous

because the numerator is $$x^2$$

27. anonymous

no $$x^3$$ in it

28. anonymous

that means $$A=0$$

29. anonymous

So the exponent in the denominator cannot be greater than x^2?

30. anonymous

look at the original question

31. anonymous

Ok

32. anonymous

or even just this part $(Ax+B)(x+2)(x-2)+C(x^2+2)(x-2)+D(x^2+2)(x+2)=x^2$

33. anonymous

on the right hand side of the equal sign, there is only a little lonely $$x^2$$ no $$x^3$$ at all

34. anonymous

on the left side there will be, when you multiply out, one $$x^3$$ term it will be $$Ax^3$$

35. anonymous

So A must equal 0 in order for the equation to be true

36. anonymous

ooh no scratch that, i am totally wrong, sorry

37. anonymous

No worries.

38. anonymous

there are other $$x^3$$ terms there i forgot about them

39. anonymous

so yeah, substitute back what you know to solve for $$A$$ and$$B$$

40. anonymous

you will find that $$A=0$$ in any case

41. anonymous

$x^2 = Ax+B(x+2)(x-2) - (1/6)(x^2+2)(x-2)+(1/6)(x^2+2)(x+2)$

42. anonymous

correct?

43. anonymous

yeah, now lets find $$B$$

44. anonymous

without doing a raft of solving a system of equations, which is the other long and tedious method of doing this

45. anonymous

$(Ax+B)(x+2)(x-2)-\frac{1}{6}(x^2+2)(x-2)+\frac{1}{6}(x^2+2)(x+2)=x^2$

46. anonymous

what will the constant be? the term without any $$x$$ or any $$x^2$$?

47. anonymous

if it is not clear what i mean, let me know

48. anonymous

Ok i will work on the problem and let you know.

49. anonymous

Ok, I am confused. I am stuck on the part where you said "what will the constant be? the term without any x"

50. anonymous

from here $-\frac{1}{6}(x^2+2)(x-2)$ when you mutiply out, you will get a number without an $$x$$ namely $-\frac{1}{6}\times 2\times (-2)=\frac{2}{3}$

51. anonymous

from here $\frac{1}{6}(x^2+2)(x+2)$ you will also get $$\frac{2}{3}$$

52. anonymous

Ok. That makes sense.

53. anonymous

from here $(Ax+B)(x+2)(x-2)$ then number will be $$4B$$

54. anonymous

]scratch that it will be $-4B$

55. anonymous

Ok makes sense.

56. anonymous

on the right hand side there is no constant at all,

57. anonymous

that means $\frac{2}{3}+\frac{2}{3}-4B=0$or $4B=\frac{4}{3}$ making $B=\frac{1}{3}$

58. anonymous

Ok. Why does the right side equal 0 though?

59. anonymous

ok i was mistaken before when i was trying to explain why $$A=0$$ but now i am not the original numerator is just $$x^2$$ it has no constant in it i

60. anonymous

the constant is 1

61. anonymous

no

62. anonymous

Sorry I mean coefficient. yes there is no constant

63. anonymous

right

64. anonymous

you got $(Ax+B)(x+2)(x-2)-\frac{1}{6}(x^2+2)(x-2)+\frac{1}{6}(x^2+2)(x+2)=x^2$ the contant on the left is $\frac{4}{3}-4B$ and on the right it is $$0$$

65. anonymous

Makes sense now.

66. anonymous

that makes $$B=\frac{1}{3}$$

67. anonymous

ok good

68. anonymous

You made the constants equal to each other.

69. anonymous

So I get the answer: [(1/3)/(x^2+2)] + [(-1/6)/(x+2)] + [(1/6)/(x-2)] ?

70. anonymous

yes sometimes you can get away without doing that, but not in this case because of the $$x^2+2$$ which is never zero so you can't use the previous trick of substitution

71. anonymous

that is one way to write it, yes

72. anonymous

Another way to write it would be [1/3(x^2+2)] - [1/6(x+2)] + [1/6(x-2)] ?

73. anonymous

So would my answer that I put be correct as well?

74. anonymous

normally you would write $\frac{1}{3(x^2+2)}-\frac{1}{6(x+2)}+\frac{1}{6(x-2)}$

75. anonymous

it is the same only in this topic do you see compound fractions allowed as an answer, but yes, both are the same

76. anonymous

Ok that makes sense. Thank you so much for you time I appreciate it. Good luck with your studies

77. anonymous

yw