Question

Write a polynomial function in standard form with the given zeros.
-1, -1, 2i

How do I do this with a imaginary number?

Answer 1

\[(x+1)(x+1)(x+2i)(x-2i)\] is what you need

Answer 2

The imaginary number should be a zero for the function. If you should solve this \[x^{2}+4=0\]
what would be the solutions?

Answer 3

if \(2i\) is a zero, then so is \(-2i\) and the quadratic will be what @suivpasyonw said, \(x^2+4\)

Answer 4

first you need to find the remaining zeros for the complex numbers.
for 2i it would be -2i

Answer 5

exactly satellite73

Answer 6

so now you got your zeros: -1, -1, 2i, -2i

Answer 7

now plug those zeros into the equation:
(x-?)(x-?)(x-?)(x-?)
(x-(-1)) (x-(-1)) (x-2i) (x-(-2i)
(x+1)(x+1)(x-2i)(x+2i)

Answer 8

i understand it now!