degree 5; zeros: -8; -I; -3+i

Question

how do I solve

Answer 1

If a complex number is a zero, then it's conjugate is also a zero...

In other words, if a+bi is a zero, then so is a-bi

 

Your zeroes are at: -8, -i, i, -3+i, -3-i

 

P(x) = (x+8)(x+i)(x-i)(x+3-i)(x+3+i)

 

Multiplying this all out, if needed, results in:

 

p(x) = x⁵ + 14x⁴ + 58x³ + 88x² + 48x + 80