# pre-cal **need help for test in the morning

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

• repeat -

• pre-cal **need help for test in the morning -

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 -

Thanks

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