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Factorization for x^3 - 4x^2 -x = 0

  1. Jul 12, 2006 #1
    for x^3 - 4x^2 -x = 0 , i have found one of the root which is 1 by dividing this equation by (x-1).
    from there onwards i can't do already to find the other two roots.somebody pls help

    thanx
     
  2. jcsd
  3. Jul 12, 2006 #2

    arildno

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    1 is not a root, 0 is. Unless you have posted the wrong equation.
     
  4. Jul 12, 2006 #3
    sorry.the matrix is [1 2 1;2 1 1;1 1 2].so i want to find the eigenvalues
     
  5. Jul 12, 2006 #4
    therefore i got the eqn x^3 - 4x^2 -x = 0
     
  6. Jul 12, 2006 #5

    matt grime

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    fine, but 1 is still not a root (1-4-1 is not zero)
     
  7. Jul 13, 2006 #6

    HallsofIvy

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    The characteristic equation for the matrix you give is x^3- 4x^2- x+ 4=0
    not what you give. That equation does have 1 as a root so apparently you just wrote the equation wrong (twice!).
    You say you found that 1 was a root "by dividing this equation by (x-1)" (Which makes me wonder why you chose x-1. It's simpler just to set x= 1 in the equation!). When you did that surely you found that
    x^3- 4x^2- x+ 4= (x-1)(x^2- 3x- 4). Solve x^2- 3x- 4= 0. That factors easily but even it it didn't you could use the quadratic formula.
     
  8. Jul 13, 2006 #7

    arildno

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    If you find the quadratic formula eerie, recognize that (1,1,1) is an eigenvector of your matrix.
     
  9. Jul 13, 2006 #8
    oooo.....okok thanx very much
     
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