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Partial fraction decompostion

  1. May 17, 2007 #1
    (2x+3)/(x+1)^2


    so this is what i am thinking...but it does not make sense


    =(A/x+1)+(B/(x+1)^2)

    so then 2x+3=A(x+1)^2+B(x+1)

    2x+3=Ax^2+A2x+A+Bx+1

    so that would make...
    0=A
    2=2A+B
    3=A+B

    this solution does not make any sense becuase if A=0 then according to the second equation B=2 which is not what B would equal in the third equation.

    what am i doing wrong?
     
  2. jcsd
  3. May 17, 2007 #2

    Dick

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    Your algebra is wrong

    2x+3=A*(x+1)+B is correct.

    I'm not sure how you are getting the equation you are working with.
     
  4. May 17, 2007 #3
    what happened to to (x+1)^2 term?
     
  5. May 17, 2007 #4

    cristo

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    This is wrong. Multipying both sides of (*) by (x+1)^2 gives 2x+3=A(x+1)+B
     
    Last edited: May 17, 2007
  6. May 17, 2007 #5

    Dick

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    Multiplying both sides by (x+1)^2 cancels one (x+1) from the A term and both from the B term.
     
  7. May 17, 2007 #6
    so should it be 2x+3/(x+1)(x+1)=(A/x+1)+(B/x+1)

    here..i just split up the (x+1)^2


    this is right?
     
  8. May 17, 2007 #7

    Dick

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    (2x+3)/(x+1)^2=A/(x+1)+B/(x+1)^2. CAREFULLY multiply each of those three terms by (x+1)^2 and report back the results.
     
  9. May 17, 2007 #8
    ohhh ok...i think i understand..

    so if i have (2x+3)/(x+1)^2=(A/x+1)+(B/(x+1)^2) (*)

    then 2x+3=A(x+1)+B cuz i multiply both sides by (x+1)^2
     
  10. May 17, 2007 #9
    thank you very much
     
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