How do you find all the zeros of f(x) = x^3 + 11x^2 + 39x + 29f(x)=x3+11x2+39x+29?

1 Answer
A. S. Adikesavan
Apr 25, 2016

x=-1. (-1+-sqrt 5)/2x=−1.−1±√52.

Explanation:

There are no changes in signs of the coefficients. So, there are no positive roots.

f(-x)=-x^3+11x^2-39x+29f(−x)=−x3+11x2−39x+29. The sum of the coefficients in f(-x)=-1+11-39+20=0. So, f(−x)=−1+11−39+20=0.So,-1 is a root and (x + 1 ) is a factor of f(x).

The other quadratic factor is readily seen as x^2+11x+29x2+11x+29. Equating this to 0 and solving, x= (-1+-sqrt 5)/2x=−1±√52..

So, the three roots are x=-1. (-1+-sqrt 5)/2.x=−1.−1±√52.

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