1-√3,1 + √3 - 2 find polynomial function of degree with given numbers as zeros

A  value of x = x1 is said to be the zero of a polynomial
f(x), if we substitute x1 for x in the polynomial f(x), then f(x) vanishes. So if f(x)
is a polynomial, and if  f(x1) = 0, then x1 is a zero of the polynomial
 f(x).

We know that if x1,x2 and x3 are the zeros of a
polynomial f(x), then the polynomial  could be written
as:

f(x) = (x-x1)(x-x2)(x-x3), as putting x= x1 or x= x2
and x=x3 makes th polynomial  vanish.

Given are the  zeros
: 1-sqrt3, 1+sqrt3 and -2 are the 3 zero.

Then f(x) = {x -
(1-sqr3)}{x-(1+sqrt3)}{x-(-2)}.

f(x) = {x^2
+(-1+sqrt3-1-sqrt3)x + (1-sqrt3)(1+sqrt3)}{x+2}.

f(x) =
{x^2 -2x + [1^2-(sqrt3)^2]}(x+2).

f(x) =
{x^2-2x+1-3}(x+2).

f(x) =
(x^2-2x-2)(x+2).

f(x) = {x^3-2x^2-2x
+2x^2-4x-4}.

f(x) =x^3 -6x -4
.

Therefore x^3-6x-4 is the polynomial with zeros 1-sqrt3 ,
1+sqrt3 ,   -2.