Zeroes of the Polynomial:

The given polynomial is 2x^4 - 9x^3 + 5x^2 + 2x - 1. We are given that two of its zeroes are 2√3 and 2-√3. Let's find the remaining zeroes using the given information.

Sum and Product of Zeroes:

The sum of zeroes of a polynomial is given by the formula:

Sum of zeroes = - (coefficient of x^3 / coefficient of x^4)

In this case, the coefficient of x^3 is -9 and the coefficient of x^4 is 2. Thus, the sum of zeroes is:

Sum of zeroes = - (-9 / 2) = 9/2

The product of zeroes of a polynomial is given by the formula:

Product of zeroes = constant term / coefficient of x^4

In this case, the constant term is -1 and the coefficient of x^4 is 2. Thus, the product of zeroes is:

Product of zeroes = -1 / 2 = -1/2

Using Sum and Product of Zeroes:

We can now form a quadratic equation using the remaining zeroes and the sum and product of zeroes.

Let the remaining zeroes be a and b. We already know that a + b = 9/2 and ab = -1/2.

The quadratic equation can be formed as (x - a)(x - b) = 0.

Expanding the equation, we get x^2 - (a + b)x + ab = 0.

Substituting the values, we have x^2 - (9/2)x - 1/2 = 0.

Solving the Quadratic Equation:

To find the zeroes of the quadratic equation x^2 - (9/2)x - 1/2 = 0, we can use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / 2a

In this case, a = 1, b = -9/2, and c = -1/2. Substituting the values, we have:

x = (9/2 ± √((-9/2)^2 - 4(1)(-1/2))) / 2(1)
x = (9/2 ± √(81/4 + 2/2)) / 2
x = (9/2 ± √(81/4 + 4/4)) / 2
x = (9/2 ± √(85/4)) / 2
x = (9/2 ± (√85)/2) / 2
x = (9 ± √85) / 4

Therefore, the remaining zeroes are (9 + √85)/4 and (9 - √85)/4.

Summary:

The zeroes of the polynomial 2x^4 - 9x^3 + 5x^2 + 2x - 1, given that two of its zeroes are 2√3 and 2-√3, are:

1. 2√3
2. 2-√3
3. (9 + √85)/4
4. (9 -