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