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

1 Answer
Apr 25, 2016

Answer:

#x=-1. (-1+-sqrt 5)/2#.

Explanation:

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

#f(-x)=-x^3+11x^2-39x+29#. The sum of the coefficients in #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+29#. Equating this to 0 and solving, #x= (-1+-sqrt 5)/2#..

So, the three roots are #x=-1. (-1+-sqrt 5)/2.#