find the zeros

p(x) = x³ -6x² + 11x - 6

On substituting x =1 in the polynomial.

p(1) = 1 - 6 + 11 - 6 = 0

So, (x-1) is a factor of the given polynomial p(x).

By long division, we get

So, x³ -6x² + 11x - 6 = (x-1) (x² -5x +6)

                                    = (x-1) (x² -2x -3x +6)

                                    = (x-1)(x-2) (x-3)

So, the zeroes of the polynomial are 1,2 and 3.