# How do you write a polynomial with zeros 8,-i, i?

Feb 11, 2016

This polynomial is $\left({x}^{3} - 8 {x}^{2} + x - 8\right)$

#### Explanation:

We can use the factored form of polynomials to write this first, then FOIL it out into the STANDARD FORM.

since the polynomial has 3 given zeroes, we can write

$\left(x - 8\right) \left(x + i\right) \left(x - i\right)$
We do this because letting x = 8,-i, or i would mean that the polynomial becomes (with an intended name) zero.

Now we can just distribute it out.

First do the two complex zeroes
$\left(x + i\right) \left(x - i\right) = {x}^{2} - {i}^{2}$
${i}^{2} = - 1$
so ${x}^{2} - {i}^{2}$ becomes ${x}^{2} + 1$
Now, just FOIL out $\left(x - 8\right) \left({x}^{2} + 1\right)$

$\left({x}^{3} - 8 {x}^{2} + x - 8\right)$
And we are done.