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Posted by on Sunday, December 12, 2010 at 11:50pm.

Find zeros:
f(x)=12x^(3)-107x^(2)-15x+24

  • repeat - , Monday, December 13, 2010 at 12:15am

    http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut38_zero1.htm#step2

  • pre-cal **need help for test in the morning - , Monday, December 13, 2010 at 12:23am

    Given:
    f(x)=12x^(3)-107x^(2)-15x+24

    Also if the leading coefficient (12) is positive, the function evaluates to -∞ at x=-∞ and to ∞ at x=∞. Thus it crosses the x axis at at least one place, or there is at least one real root.

    In fact, application of Descarte's rule indicate that there are two changes of sign, so there are at least two positive roots. This also means that all three roots are real, since complex roots must come in pairs (conjugates).

    Equate
    f'(x)=36x²-214x-15
    to zero, and solve for the maximum and minimum, if any.
    f'(x)=0 at x=-0.07 and x=6.01
    f(-0.07)=24.52 (maximum)
    f(6.01)=-1326,
    so there is a root close to 1 and probably another around -1.

    By numerical methods, such as Newton's method, it should be possible to evaluate the zeroes as:
    x=-0.531, x=0.417, and x=9.031

    Check:
    (x1*x2*x3)=2=24/12
    (x1+x2+x3)=8.917 = 107/12

  • pre-cal **need help for test in the morning - , Monday, December 13, 2010 at 12:25am

    Thanks

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