Write the original polynomial given zeros of -3 and -5i

Question

x= -3 x=-5i x=5i 

(x-3) (x-5i) (x+5i) 

(x-3) (x^2+5ix-5ix-25i^2)

i got suck after that 

Answer 1

It must be of minimum degree 3.  Assuming a leading coefficient of 1, and knowing that the complex conjugate of a complex zero is also a zero (in order to have real coefficients),

 

p(x) = [x - (-3)] [x - (-5i)] (x - 5i)

 

p(x) = (x + 3)(x + 5i)(x - 5i)

 

Note that

 

(x + 5i)(x - 5i) = x² - (5i)² = x² + 25

 

Then

 

p(x) = (x + 3)(x² + 25)

 

You should have no trouble finishing from here.