# How do you find all the zeros of F(x)=x^3-3x^2+9x+13?

Mar 5, 2016

Identify $x = - 1$ as a zero by examining the coefficients, separate out the corresponding factor $\left(x + 1\right)$, then solve the remaining quadratic factor by completing the square.

#### Explanation:

First notice that reversing the signs of the coefficients of the terms with odd degree results in coefficients that sum to zero.

That is $- 1 - 3 - 9 + 13 = 0$.

So $F \left(- 1\right) = 0$ and $\left(x + 1\right)$ is a factor:

${x}^{3} - 3 {x}^{2} + 9 x + 13$

$= \left(x + 1\right) \left({x}^{2} - 4 x + 13\right)$

$= \left(x + 1\right) \left({x}^{2} - 4 x + 4 + 9\right)$

$= \left(x + 1\right) \left({\left(x - 2\right)}^{2} - {\left(3 i\right)}^{2}\right)$

$= \left(x + 1\right) \left(\left(x - 2\right) - 3 i\right) \left(\left(x - 2\right) + 3 i\right)$

$= \left(x + 1\right) \left(x - 2 - 3 i\right) \left(x - 2 + 3 i\right)$

So the zeros of $F \left(x\right)$ are $x = - 1$ and $x = 2 \pm 3 i$