F( x ) = ( x - 1 ) ( x + 2i )( x - 2i ), because the zeroes are 1, -2i, +2i ;
f( x ) = ( x - 1 )( x^2 - 4i^2 ) = ( x - 1 )( x^2 + 4 ) = x^3 - x^2 + 4x - 4 ;
Answer:
3,3 and 3,30
Step-by-step explanation:
No matter how many zeros you wax on or wane off, you will still have equivalent values.
I am joyous to assist you anytime.
Are you asking like thousands or something else?
A) (-4,2π/3)
b) (4,4π/3)
c) (-2,π/3)
d) (2,5π/3)
I believe that would be 1
