# How do you write a polynomial with zeros 2i, 3i, 1?

Apr 24, 2016

${x}^{5} - {x}^{4} + 13 {x}^{3} - 13 {x}^{2} + 36 x - 36 = 0$

#### Explanation:

Complex roots occur in conjugate pairs. So, the roots are $1 , \pm 2 i , \pm 3 i ,$. The degree of the polynomial is 5.

Hence, the polynomial equation is

$\left(x - 1\right) \left(\left(x - 2 i\right) \left(x + 2 i\right)\right) \left(\left(x - 3 i\right) \left(x + 3 i\right)\right) = 0$

$\left(x - 1\right) \left({x}^{2} + {2}^{2}\right) \left({x}^{2} + {3}^{2}\right) = 0$

Expanding,

${x}^{5} - {x}^{4} + 13 {x}^{3} - 13 {x}^{2} + 36 x - 36 = 0$