11x^3-99x^2+315=0

find the roots of the equation
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1 Answer

I think you may mean 11x3-99x2-35x+315=11x2(x-9)-35(x-9)=(11x2-35)(x-9).

So 11x2=35, x2=35/11, x=±√(35/11); or x=9.

The three zeroes are 9, 1.7838 and -1.7838 (approx).

by Top Rated User (1.2m points)

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