Complex solution of a cubic function.

[yoyoyo992]

Hi,
I'm studying my algebra and I was trying this exercise:

Solve
z³+iz²-7z-iz-6-6i=0

I found the 2 real solutions (3 and -2) but i can't seem to find the complex one.

I tried this:

(z²-z)(i)=6(i) ==> z= -2 and 3
z³-7z=6==> z= -1, -2 and 3

I found that the complex solution is -1-i but i have no idea how to find it.

(Im from Belgium, it's difficult to type maths in another language so sorry for mistakes)

 

[HallsofIvy]

Well, the obvious thing to do is divide by z+ 2 and z- 3 so that you can factor that cubic into linear terms, then set each term equal to 0.

 

[yoyoyo992]

Thanks, I found it