# How do you write a polynomial with zeros 3i, -3i, 5 and leading coefficient 1?

Nov 17, 2016

#### Explanation:

y = 1

Multiply by the factor corresponding to the root $3 i$:

$y = \left(x - 3 i\right)$

Multiply by the factor corresponding to the root $- 3 i$:

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

Multiply by the factor corresponding to the root 5:

$y = \left(x - 3 i\right) \left(x + 3 i\right) \left(x - 5\right)$

When we multiply an complex conjugate pair $\left(a \pm b\right)$, we know that we get the sum of 2 squares $\left({a}^{2} + {b}^{2}\right)$:

$y = \left({x}^{2} + 9\right) \left(x - 5\right)$

Use the F.O.I.L. method:

$y = {x}^{3} - 5 {x}^{2} + 9 x - 45$