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Finding x intercept

  1. Dec 1, 2005 #1
    Hey, Is there any simple way of finding the x intercept for

    x^3-6x^2-15x+4
     
  2. jcsd
  3. Dec 1, 2005 #2
    Basically find the roots of the equation?
    I bet you already know that there are three roots.
    to solve for x in a cubic of form ax^3+bx^2+cx+d, use a program like mathematica, or check out this cubic forumla:

    http://www.math.vanderbilt.edu/~schectex/courses/cubic/cubic.gif

    plug in 1 for a, 6 for b, 15 for c, and 4 for d.
    pretty long, but it always works.
     
    Last edited: Dec 1, 2005
  4. Dec 2, 2005 #3

    HallsofIvy

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    As above, the only way to find the x-intercepts is to set y= 0 and solve the equation. This equation has no rational roots so there is no "simple" way to do it.
     
  5. Dec 3, 2005 #4
    About newtowns method?
     
  6. Dec 3, 2005 #5
    what about it? [tex] x_{n+1} = x_n - \frac {f(x_n)}/{f'(x_n)} [/tex]
    find the derivative, and start solving...


    edit:

    there's supposed to be a "x_n - "before the fraction, but it isnt showing up...
     
  7. Dec 3, 2005 #6
    Thats another way of finding the x intercepts? Right?
    (Our teacher never mentioned it, i think im gonna have to learn it myself)
     
  8. Dec 3, 2005 #7

    HallsofIvy

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    Finding the x-intercepts for
    x is exactly the same as finding the solutions to
    x^3-6x^2-15x+4= 0. There is no "trivial?" way of doing that.
     
  9. Dec 19, 2005 #8
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