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Solve cubic with no rational zeros

  1. Sep 18, 2012 #1
    1. The problem statement, all variables and given/known data

    x^3-8x+10=0

    2. Relevant equations



    3. The attempt at a solution

    By dividing by x-1 (1 is a factor of the solution), I got (x-1)(x2+x-7)+3+10=0

    which equals x^3-8x+20=0

    I think the solution would be imaginary but I used a graphing calculator and there is one solution but I don't know how to calculate it.
     
    Last edited by a moderator: Sep 19, 2012
  2. jcsd
  3. Sep 19, 2012 #2

    Mark44

    Staff: Mentor

    What is the complete problem statement?
    x - 1 is NOT a factor of x^3 - 8x + 10.
    We can't help you if we don't know what it is that you're supposed to do. That's where the complete problem statement would be useful information.
     
  4. Sep 20, 2012 #3

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Unless you know how to solve a cubic by radicals, you won't be finding the exact answer for the real root. Maple gives it as:

    -(1/3)*(135+3*sqrt(489))^(1/3) - 8/(135+3*sqrt(489))^(1/3)

    and a decimal approximation to that is -3.318628218.
     
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