# Question #5eda4

Sep 4, 2017

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

#### Explanation:

The complex conjugate root theorem states that if $a + b i$ is a root of a polynomial, then so is $a - b i$. So, since $i$ is a root, then $- i$ will also be a root.

Since the polynomial has the zeros $3 , i ,$ and $- i$, the factors of the polynomial will be $x - 3$, $x - i$, and $x - \left(- i\right) = x + i$.

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

Since $\left(a + b\right) \left(a - b\right) = {a}^{2} - {b}^{2}$, we can rewrite the function as

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

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

We can expand this further.

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

In standard form, the function is $f \left(x\right) = {x}^{3} - 3 {x}^{2} + x - 3$.