Question

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

Answer 1

`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.`