# How do you find all the zeros of x^3 + x^2 + 9x + 9  given zero 3i?

May 21, 2016

#### Answer:

The zeros are $- 1$, $3 i$ and $- 3 i$.

#### Explanation:

We don't really need to be told that $3 i$ is a zero, except that it does inform us that we should include Complex zeros in the answer.

This cubic factors by grouping:

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

$= \left({x}^{3} + {x}^{2}\right) + \left(9 x + 9\right)$

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

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

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

So the zeros are $- 1$, $3 i$ and $- 3 i$.