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Polynomial divisjon

  1. Dec 2, 2006 #1
    (x^3+x^2+x-1)/(x^2+2x+2)

    I want to compute the integral of this, but in order to use the method of partial fractions, I need to transform it into something with a higher degree in the divisor. How to I compute this?

    It's supposed to be (x-1) + (x+1)/(x^2+2x+2)

    Hope someone has got a clue :smile:
     
  2. jcsd
  3. Dec 2, 2006 #2
    [tex] \int \frac{x^{3}+x^{2}+x-1}{x^{2}+2x+2} [/tex]


    Doing the long division you get [tex] x-1 [/tex] with a remainder of [tex] \frac{x+1}{x^{2}+2x+2} [/tex]
     
  4. Dec 2, 2006 #3

    radou

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    Not that I don't want to help, but polynomial division is pretty much of an 'algorythmic' procedure which can easily be found by google-ing. :wink:
     
  5. Dec 3, 2006 #4

    HallsofIvy

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    Basically, it is just the same as long division of numbers- I'm sure you learned it long ago and just need a reminder.

    Look at the leading terms: [itex]x^2[/itex] divides into [itex]x^3[/itex] x times. The first term of the quotient is x. Multiply the entire divisor by x: [/itex]x^3+ 2x^2+ 2x[/itex] and subtract: [itex]x^3+ x^2+ x- 1- (x^3+ 2x^2- 2x)= -x^2- x- 1[/itex]. Now [itex]x^2[/itex] divides into [itex]-x^2[/itex] -1 times: quotient is now x-1. Multiply the divisor by -1 to get -x^2-2x-2 and subtract: -x^2- x- 1-(-x^2- 2x-2)= x+ 1. Since that has lower degree than the divisor you are done: the quotient is x- 1 with remainder x+1.
    [tex]\frac{x^3+ x^2+ x- 1}{x^2+ 2x+ 2}= x- 1+ \frac{x+1}{x^2+ 2x+ 2}[/tex]


    NOW, do the hard part!
     
  6. Dec 3, 2006 #5
    I found this algorithm at wikipedia http://en.wikipedia.org/wiki/Polynomial_long_divisio

    The only thing I don't understand is why you don't pull down (-42) in step 3. -42-0 =-42...
     
  7. Dec 3, 2006 #6
    you can if you want to
     
  8. Dec 3, 2006 #7
    OK, that makes sense.

    Thank you all!
     
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