algebra
posted by lee .
what are the real or imaginary solutios of the polynomial equation
x^38=0
I came up with 1+i sgrt 3 and 1i sgrt 3
could some see if I'm right

First byou must find real solution.
x ^ 3  8 = 0 Add 8 to both sides
x ^ 3  8 + 8 = 0 + 8
x ^ 3 = 8
x = third root of 8 = 2
In this case one rational zero is x = 2
Now you divide polynomial x ^ 3  8 with ( x  2 )
( x ^ 3  8 ) / ( x  2 ) = x ^ 2 + 2 x + 4
OR
( x ^ 3  8 ) = ( x ^ 2 + 2 x + 4 ) * ( x  2 ) = 0
Youe equation have roots when :
x ^ 2 + 2 x + 4 = 0
AND
x  2 = 0
Solutions of equation x ^ 2 + 2 x + 4 = 0
are
1 + i sgrt 3 and 1  i sgrt 3
Soluton of equation x  2 = 0
are
x = 2
Your equation have 3 solutions.
One real solution x = 2
and two imaginary solutions :
1 + i sgrt 3 and 1  i sgrt 3 
thank you