SOLUTION: F(x)= x^3-5x^2+144x-720 I got one of the zeroes (x=5) I need the other 2... they should be imaginary numbers.

Algebra ->  Equations -> SOLUTION: F(x)= x^3-5x^2+144x-720 I got one of the zeroes (x=5) I need the other 2... they should be imaginary numbers.      Log On


   

Question 254900: F(x)= x^3-5x^2+144x-720
I got one of the zeroes (x=5) I need the other 2... they should be imaginary numbers.
Found 2 solutions by scott8148, edjones:
Answer by scott8148(6628) About Me  (Show Source):
You can put this solution on YOUR website!
if r is a root (zero) then x-r is a factor

dividing by x-5 ___ x^2 + 144

x^2 + 144 = 0 ___ x^2 = -144 ___ x = ±12i

Answer by edjones(8007) About Me  (Show Source):
You can put this solution on YOUR website!
x=5
x-5=0
(x^3-5x^2+144x-720)/(x-5) Long division
=x^2+144
x^2+144=0
x^2=-144
x=+-sqrt(-144)
=+-12i the other 2 zeros
.
Ed