Answer:
_2, 2, 3
Step-by-step explanation:
[tex]g(x) = {x}^{3} - 3 {x}^{2} - 4x + 12 \\ = {x}^{2} (x - 3) - 4(x - 3) \\ = (x - 3)( {x}^{2} - 4) \\ = (x - 3)(x - 2)(x + 2) \\ plug \: g(x) = 0 \\ \therefore \: (x - 3)(x - 2)(x + 2) = 0 \\ (x - 3) = 0 \: or \: (x - 2) = 0 \: \\ or \: (x + 2) = 0 \\ \therefore \:x = 3 \: or \: x = 2 \: or \: x = - 2 \\ \therefore \:x = \{ - 2, \: \: 2, \: \: 3 \}[/tex]
Hence, - 2, 2 & 3 are the zeros of the given function.
