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anonymous

  • one year ago

Partial Fraction decomposition: (x^2)/(x^4-2x^2-8)

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  1. anonymous
    • one year ago
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    can you factor the denominator?

  2. anonymous
    • one year ago
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    Yes i factored denominator: (x^2+2)(x-2)(x+2)

  3. anonymous
    • one year ago
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    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
    • one year ago
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    actually i think in this case \(A=0\) but it doesn't need to be

  5. anonymous
    • one year ago
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    would (A/x-2) + (B/x+2) + ((Cx+D)/(x^2+2)) be correct as well?

  6. anonymous
    • one year ago
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    now add \[Ax+B(x+2)(x-2)+C(x^2+2)(x-2)+D(x^2+2)(x+2)=x^2\]

  7. anonymous
    • one year ago
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    yes of course

  8. anonymous
    • one year ago
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    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
    • one year ago
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    Yes that's completely fine. thank you

  10. anonymous
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    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 year ago
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    one minute i am reading your work

  14. anonymous
    • one year ago
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    ok take your time i am going to get a snack if it is not clear, ask

  15. anonymous
    • one year ago
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    I understand what you did. I worked the problem and got D=1/6

  16. anonymous
    • one year ago
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    ok good

  17. anonymous
    • one year ago
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    now we can use the same gimmick only replace \(x\) yb \(-2\)

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

  19. anonymous
    • one year ago
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    yes

  20. anonymous
    • one year ago
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    and now I plugin C and D into the first equation to solve for A & B?

  21. anonymous
    • one year ago
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    you can do that, but first lets think

  22. anonymous
    • one year ago
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    Ok..

  23. anonymous
    • one year ago
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    the first part is \[(Ax+B)(x+2)(x-2)\]

  24. anonymous
    • one year ago
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    if you multiply that out mentally you get an \(Ax^3\) term there is not \(x^3\) so \(A=0\)

  25. anonymous
    • one year ago
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    Sorry I do not understand why Ax^3 = 0

  26. anonymous
    • one year ago
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    because the numerator is \(x^2\)

  27. anonymous
    • one year ago
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    no \(x^3\) in it

  28. anonymous
    • one year ago
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    that means \(A=0\)

  29. anonymous
    • one year ago
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    So the exponent in the denominator cannot be greater than x^2?

  30. anonymous
    • one year ago
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    look at the original question

  31. anonymous
    • one year ago
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    Ok

  32. anonymous
    • one year ago
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    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
    • one year ago
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    on the right hand side of the equal sign, there is only a little lonely \(x^2\) no \(x^3\) at all

  34. anonymous
    • one year ago
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    on the left side there will be, when you multiply out, one \(x^3\) term it will be \(Ax^3\)

  35. anonymous
    • one year ago
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    So A must equal 0 in order for the equation to be true

  36. anonymous
    • one year ago
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    ooh no scratch that, i am totally wrong, sorry

  37. anonymous
    • one year ago
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    No worries.

  38. anonymous
    • one year ago
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    there are other \(x^3\) terms there i forgot about them

  39. anonymous
    • one year ago
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    so yeah, substitute back what you know to solve for \(A\) and\(B\)

  40. anonymous
    • one year ago
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    you will find that \(A=0\) in any case

  41. anonymous
    • one year ago
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    \[x^2 = Ax+B(x+2)(x-2) - (1/6)(x^2+2)(x-2)+(1/6)(x^2+2)(x+2)\]

  42. anonymous
    • one year ago
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    correct?

  43. anonymous
    • one year ago
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    yeah, now lets find \(B\)

  44. anonymous
    • one year ago
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    without doing a raft of solving a system of equations, which is the other long and tedious method of doing this

  45. anonymous
    • one year ago
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    \[(Ax+B)(x+2)(x-2)-\frac{1}{6}(x^2+2)(x-2)+\frac{1}{6}(x^2+2)(x+2)=x^2\]

  46. anonymous
    • one year ago
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    what will the constant be? the term without any \(x\) or any \(x^2\)?

  47. anonymous
    • one year ago
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    if it is not clear what i mean, let me know

  48. anonymous
    • one year ago
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    Ok i will work on the problem and let you know.

  49. anonymous
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    from here \[\frac{1}{6}(x^2+2)(x+2)\] you will also get \(\frac{2}{3}\)

  52. anonymous
    • one year ago
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    Ok. That makes sense.

  53. anonymous
    • one year ago
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    from here \[(Ax+B)(x+2)(x-2)\] then number will be \(4B\)

  54. anonymous
    • one year ago
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    ]scratch that it will be \[-4B\]

  55. anonymous
    • one year ago
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    Ok makes sense.

  56. anonymous
    • one year ago
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    on the right hand side there is no constant at all,

  57. anonymous
    • one year ago
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    that means \[\frac{2}{3}+\frac{2}{3}-4B=0\]or \[4B=\frac{4}{3}\] making \[B=\frac{1}{3}\]

  58. anonymous
    • one year ago
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    Ok. Why does the right side equal 0 though?

  59. anonymous
    • one year ago
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    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
    • one year ago
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    the constant is 1

  61. anonymous
    • one year ago
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    no

  62. anonymous
    • one year ago
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    Sorry I mean coefficient. yes there is no constant

  63. anonymous
    • one year ago
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    right

  64. anonymous
    • one year ago
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    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
    • one year ago
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    Makes sense now.

  66. anonymous
    • one year ago
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    that makes \(B=\frac{1}{3}\)

  67. anonymous
    • one year ago
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    ok good

  68. anonymous
    • one year ago
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    You made the constants equal to each other.

  69. anonymous
    • one year ago
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    So I get the answer: [(1/3)/(x^2+2)] + [(-1/6)/(x+2)] + [(1/6)/(x-2)] ?

  70. anonymous
    • one year ago
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    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
    • one year ago
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    that is one way to write it, yes

  72. anonymous
    • one year ago
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    Another way to write it would be [1/3(x^2+2)] - [1/6(x+2)] + [1/6(x-2)] ?

  73. anonymous
    • one year ago
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    So would my answer that I put be correct as well?

  74. anonymous
    • one year ago
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    normally you would write \[\frac{1}{3(x^2+2)}-\frac{1}{6(x+2)}+\frac{1}{6(x-2)}\]

  75. anonymous
    • one year ago
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    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
    • one year ago
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    Ok that makes sense. Thank you so much for you time I appreciate it. Good luck with your studies

  77. anonymous
    • one year ago
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    yw

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