What are the zeros of (x-2)(x^2-9)

So because the factors of the equation are (x-2), (x+3) and (x-3), you set them equal to zero and you get x=2, -3, and 3

                         (x - 2)(x² - 9) = 0
          (x - 2)(x² + 3x - 3x - 9) = 0
(x - 2)(x(x) + x(3) - 3(x) - 3(3)) = 0
                 (x - 2)(x - 3)(x + 3) = 0
x - 2 = 0  or  x - 3 = 0  or  x + 3 = 0
  + 2 + 2         + 3  + 3          - 3  - 3
     x = 2             x = 3             x = -3

If i’m not wrong i believe it is c