# How do you write a polynomial function of least degree given the zeros 2i, -2i, 2+2i?

Oct 14, 2016

Please see the explanation for the process.

$y = k \left({x}^{4} - 4 {x}^{3} + 12 {x}^{2} - 16 x + 32\right)$

#### Explanation:

You cannot have an odd number of complex or imaginary roots, because they always exist is conjugate pairs, therefore, another root must be 2 - 2i and the polynomial is of the form:

$y = k \left(x - 2 i\right) \left(x + 2 i\right) \left(x - 2 - 2 i\right) \left(x - 2 + 2 i\right)$

$y = k \left({x}^{2} - 4 {i}^{2}\right) \left(x - 2 - 2 i\right) \left(x - 2 + 2 i\right)$

$y = k \left({x}^{2} + 4\right) \left(x - 2 - 2 i\right) \left(x - 2 + 2 i\right)$

$y = k \left({x}^{2} + 4\right) \left({x}^{2} - 2 x + 2 i x - 2 x + 4 - 4 i - 2 i x + 4 i - 4 {i}^{2}\right)$

$y = k \left({x}^{2} + 4\right) \left({x}^{2} - 4 x + 4 - 4 {i}^{2}\right)$

$y = k \left({x}^{2} + 4\right) \left({x}^{2} - 4 x + 4 + 4\right)$

$y = k \left({x}^{2} + 4\right) \left({x}^{2} - 4 x + 8\right)$

$y = k \left({x}^{4} - 4 {x}^{3} + 8 {x}^{2} + 4 {x}^{2} - 16 x + 32\right)$

$y = k \left({x}^{4} - 4 {x}^{3} + 12 {x}^{2} - 16 x + 32\right)$

k exists to allow the polynomial to pass through a specified point.