Question

what are the real or imaginary solutios of the polynomial equation
x^3-8=0

I came up with 1+i sgrt 3 and 1-i sgrt 3
could some see if I'm right

Answer 1

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

Answer 2

thank you