# How do you write a polynomial with zeros 8, -i, i and leading coefficient 1?

Mar 2, 2017

$\textcolor{g r e e n}{{x}^{3} - 8 {x}^{2} + x - 8}$

#### Explanation:

If the polynomial has zeros $8 , - i , \mathmr{and} i$
then it has factors $c \times \left(x - 8\right) \left(x + i\right) \left(x - i\right)$ where $c$ is a constant.

Since we are asked for a polynomial with a leading coefficient of $1$,
$c = 1$ and we can ignore it from here on.

$\textcolor{b l u e}{\left(x + i\right) \left(x - i\right)} = {x}^{2} - {i}^{2} = {x}^{2} - \left(- 1\right) = \textcolor{b l u e}{{x}^{2} + 1}$
and
(x-8)color(blue)((x+i)(x-i))=(x-8)xxcolor(blue)(""(x^2+1))=x^3-8x^2+x-8