How can i find the number of complex, real,and imaginary roots?

x = 2 || x = -1 - I Sqrt[3] || x = -1 + I Sqrt[3]


There is 3 roots


The imaginary roots must be is pairs because the coefficients are real

solve(x^3 - 8  =  0)

x1 = 2, x2 = - 1 - I*sqrt(3), x3 = - 1 + I*sqrt(3)

the number of roots (imaginary and real) is = to the highest degree yes.
first you want to factor
(x-2)(x^2+2x+4)
so x=2 and for this part x^2+2x+4 you just plug it into the quadratic formula you get -2 plus or minus sqroot of 4-4(1)(4) divided by 2(1) then you get -2 plus or minus squareroot of -12 all over 2 you replace the negative with i (b/c negative square root is imaginary) you are left with -2plus or minus 2isquare root of 3 all divided by 2 cancel the two's you have -1plus or minus isquare root of 3, and the real root from the beginning 2.