Posted by Terri on Sunday, December 12, 2010 at 11:50pm.
Find zeros:
f(x)=12x^(3)107x^(2)15x+24

repeat  TutorCat, Monday, December 13, 2010 at 12:15am
http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut38_zero1.htm#step2

precal **need help for test in the morning  MathMate, 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)=36x²214x15
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 
precal **need help for test in the morning  Terri, Monday, December 13, 2010 at 12:25am
Thanks