Solution:

Given polynomial is 2x^4 - 13x^3 + 19x^2 + 7x - 3.

Two zeroes of the given polynomial are (2 + √3) and (2 - √3).

Let us find the other zeroes of the polynomial.

By factor theorem, if (2 + √3) is a zero of the given polynomial, then (x - 2 - √3) is a factor of the polynomial.

Similarly, if (2 - √3) is a zero of the given polynomial, then (x - 2 + √3) is a factor of the polynomial.

Now, we can write the polynomial as a product of linear factors as follows:

2x^4 - 13x^3 + 19x^2 + 7x - 3 = 2(x - 2 - √3)(x - 2 + √3)(ax^2 + bx + c)

where a, b, c are constants to be determined.

Expanding the above expression, we get:

2x^4 - 13x^3 + 19x^2 + 7x - 3 = 2(x^2 - 4x + 1)(ax^2 + bx + c)

Multiplying the brackets, we get:

2x^4 - 13x^3 + 19x^2 + 7x - 3 = 2ax^4 + (2bx^3 - 8ax^3) + (2cx^2 - 8bx^2 + x^2) + (-4cx + 7x) - 3c

Equating the coefficients of the like terms on both sides, we get:

2a = 2 => a = 1

2b - 8a = -13 => b = -3

2c - 8b + 1 = 19 => c = 3

-4c + 7 = 7 => c = 0

-3c - 3 = -3 => c = 1

Therefore, the polynomial can be written as:

2x^4 - 13x^3 + 19x^2 + 7x - 3 = 2(x - 2 - √3)(x - 2 + √3)(x^2 - 3x + 1)

The other two zeroes of the polynomial can be found by solving the quadratic factor x^2 - 3x + 1.

Using the quadratic formula, we get:

x = [3 ± √(9 - 4)]/2 = [3 ± √5]/2

Therefore, the four zeroes of the polynomial are:

2 + √3, 2 - √3, [3 + √5]/2, [3 - √5]/2

Hence, the solution.