Find all zeros of

f (x) = x^3 + 3x^2 - 3x - 1

F (1) = 1^3 + 3 (1) ^1 - 3 - 1 = 1 + 3 - 3 - 1 = 0

so one zero is x = 1, and x-1 is a factor

dividing f (x) by x - 1 gives x^2 + 4x + 1

x = [-4 + / - sqrt (16 - 4) ] / 2

x = - 0.27, - 3.73

zeroes are 1, - 0.27 and - 3.73 Answer