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anonymous

  • 5 years ago

Find the partial fraction decomposition of the rational function. 9x^2 − 3x − 9 x^4 + 3x3

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  1. anonymous
    • 5 years ago
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    are you looking for an answer or the process?

  2. anonymous
    • 5 years ago
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    both

  3. anonymous
    • 5 years ago
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    well first your need to set up the equivalence \[\frac{9x^2-3x-9}{x^3(x+3)}=\]\[\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x^3}+\frac{D}{x+3}\]

  4. anonymous
    • 5 years ago
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    ok

  5. anonymous
    • 5 years ago
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    Thats whats confusing me

  6. anonymous
    • 5 years ago
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    what is confusing me is that this interface is buggy

  7. anonymous
    • 5 years ago
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    now it works...

  8. anonymous
    • 5 years ago
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    yea

  9. anonymous
    • 5 years ago
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    then rewrite that as \[9x^2-3x-0=\]\[Ax^2(x+3)+Bx(x+3)\]\[+C(x+3)+Dx^3\]

  10. anonymous
    • 5 years ago
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    that should be a 9 where that 0 is

  11. anonymous
    • 5 years ago
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    k

  12. anonymous
    • 5 years ago
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    then rewrite that as \[9x^2-3x-9=\]\[Ax^2(x+3)+Bx(x+3)\]\[+C(x+3)+Dx^3\]

  13. anonymous
    • 5 years ago
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    so are we good to this point?

  14. anonymous
    • 5 years ago
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    yes

  15. anonymous
    • 5 years ago
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    now we can choose x=0 and determine that C=-3

  16. anonymous
    • 5 years ago
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    ok

  17. anonymous
    • 5 years ago
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    and next let x=-3 to determine that D=-3

  18. anonymous
    • 5 years ago
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    are you good with what I am doing?

  19. anonymous
    • 5 years ago
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    yea i get it

  20. anonymous
    • 5 years ago
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    now however we are going to need to choose two different values of x and get a system of two equations in two variable tosolve for A and B

  21. anonymous
    • 5 years ago
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    alright

  22. anonymous
    • 5 years ago
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    so I chose x=1 and got 3=A+B

  23. anonymous
    • 5 years ago
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    Ok

  24. anonymous
    • 5 years ago
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    next I chose x=-1 and got 3=A-B

  25. anonymous
    • 5 years ago
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    adding those equations together I get 6=2A which says A=3

  26. anonymous
    • 5 years ago
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    then using A=3 in either of the other two either of the previous two equations I determine that B=0

  27. anonymous
    • 5 years ago
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    Now finally substituting those values for A,B,C,D we get\[\frac{9x^2-3x-3}{x^3(x+3)}=\]\[\frac{3}{x}-\frac{3}{x^3}-\frac{3}{x+3}\]

  28. anonymous
    • 5 years ago
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    ok i see

  29. anonymous
    • 5 years ago
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    Thats alot!

  30. anonymous
    • 5 years ago
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    your welcome, glad you followed it. Long process...

  31. anonymous
    • 5 years ago
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    oh yea

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