Find the x-intercepts and y-intercept of y=x^3-2x^2-4x+8

For the intersection with the x axis we have:x ^ 3-2x ^ 2-4x + 8 = 0Rewriting we have:
 ((x-2) ^ 2) (x + 2) = 0
 The solutions are:
 x1 = 2
 x2 = -2
 The intersections are:
 (-2,0)
 (2,0)
 For the intersection with the axis and we have:
 y = x ^ 3-2x ^ 2-4x + 8
 y = (0) ^ 3-2 (0) ^ 2-4 (0) +8
 y = 8
 The point of intersection is:
 (0, 8)
 
 The x-intercepts and y-intercept of y = x ^ 3-2x ^ 2-4x + 8 are:
 (-2, 0) and (2, 0)
 (0, 8)
 respectively

Ithink it is 7. 16