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Solving fourth degree polynomial

  1. May 1, 2007 #1
    Here it is

    (x^4)-(3x^3)-(3x^2)+2 = 0

    Two solutions are

    x = 1 - i
    x = 1 + i

    How can i find the other solutions, (not graphically)

    Thanks
     
  2. jcsd
  3. May 1, 2007 #2
    General solution:
    http://en.wikipedia.org/wiki/Quartic_equation

    Numerical method:
    http://en.wikipedia.org/wiki/Newton's_method
     
  4. May 1, 2007 #3

    uart

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    There's no need for that. OP already has two of the roots so all that's required is to divide the original quartic by (x^2 - 2x + 2) and factorize the resultant quadratic.

    BTW. (x^2 - 2x + 2) = (x-(1+i)) (x-(1-i))
     
  5. May 1, 2007 #4
    good call.
     
  6. May 2, 2007 #5

    HallsofIvy

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    One serious problem the OP has is that 1-i and 1+ i are NOT roots of
    x4-3x3-3x2+2= 0!

    If x= 1- i then x4-3x3-3x2+2= 4+ 12i and if x= 1+ i then x4-3x3-3x2+2= 4- 12i, not 0.
     
  7. May 2, 2007 #6

    uart

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    Science Advisor

    Oh yeah you're right. I just took his word for that and didn't check it.

    My guess is that OP was given the two roots as part of the problem and has made a typo error in posting the equation.
     
  8. May 2, 2007 #7
    yes sorry it was a typo but i got the method

    thanks
     
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