# How do you find all the zeros of 3x^3-4x^2+8x+8 given the zero: 1-sqrt 3i?

Apr 19, 2016

$x = - \frac{2}{3} \mathmr{and} \left(1 \pm i \sqrt{3}\right)$.

#### Explanation:

Complex roots occur in conjugate pairs.
So, two of the roots are $1 \pm i \sqrt{3}$.
The sum of these roots = 2..

The sum of the three roots is
$-$(coefficient of ${x}^{2}$)/(coefficient of x^3)=$\frac{4}{3}$.

So, the third root = $\frac{4}{3} - 2 = - \frac{2}{3}$..