how do I find the equation with the following zeros: -1, 4-5i

Question

anonymous · Answer

multiply out (expand) the left side if you want standard form.

anonymous · Answer

Where did the 3 come from? this is the answer I came up with x^3-7x^2+33x+41       for the first part that you show I came up with (x+1)(x-4+5i)(x-4-5i). Is this correct?

anonymous · Answer

I have no idea!!!  it's supposed to be the -1.

anonymous · Answer

since you list a complex root, it's conjugate is also a root so you actually have 3 roots:
x = -1, x = 4-5i, x = 4+5i (conjugate)

anonymous · Answer

lol I was freaking out over here!

anonymous · Answer

so now just create your equation with the listed as your solution:
(x--1)*(x-(4-5i))(x+(4+5i)) = 0

anonymous · Answer

sorry, bout that!

anonymous · Answer

no problem, I don't fully grasp all of the steps, but I think I'm doing most of it correctly, I just don't know where I am going wrong.

anonymous · Answer

if you know the zeros to a function, always write it out in factored form then expand it.

anonymous · Answer

lost internet connection for a while....

anonymous · Answer

so you should get:  x^3 - 7x^2 + 33x + 41 = 0

anonymous · Answer

my screen froze

anonymous · Answer

ok, I stumbled through it but that's what I came up with, it's just confusing.

anonymous · Answer

do I need to put the =0 part?