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

Apr 25, 2016

$x = - 1. \frac{- 1 \pm \sqrt{5}}{2}$.

#### Explanation:

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

$f \left(- x\right) = - {x}^{3} + 11 {x}^{2} - 39 x + 29$. The sum of the coefficients in $f \left(- x\right) = - 1 + 11 - 39 + 20 = 0. S o ,$-1 is a root and (x + 1 ) is a factor of f(x).

The other quadratic factor is readily seen as ${x}^{2} + 11 x + 29$. Equating this to 0 and solving, $x = \frac{- 1 \pm \sqrt{5}}{2}$..

So, the three roots are $x = - 1. \frac{- 1 \pm \sqrt{5}}{2.}$