**Finding the Zeroes of a Quadratic Polynomial**

To find the zeroes of the quadratic polynomial x^2 + (1/6)x - 2, we need to solve the equation x^2 + (1/6)x - 2 = 0. This can be done by using the quadratic formula or by factoring the equation if possible.

**Using the Quadratic Formula**

The quadratic formula states that for any quadratic equation in the form ax^2 + bx + c = 0, the zeroes can be found using the formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

In our case, a = 1, b = 1/6, and c = -2. Plugging these values into the quadratic formula, we get:

x = (-(1/6) ± √((1/6)^2 - 4(1)(-2))) / (2(1))

Simplifying further:

x = (-1/6 ± √(1/36 + 8)) / 2

x = (-1/6 ± √(1/36 + 288/36)) / 2

x = (-1/6 ± √(289/36)) / 2

x = (-1/6 ± (17/6)) / 2

This gives us two possible values for x:

x = (-1/6 + 17/6) / 2 = 16/12 = 4/3

x = (-1/6 - 17/6) / 2 = -18/12 = -3/2

Therefore, the zeroes of the quadratic polynomial x^2 + (1/6)x - 2 are x = 4/3 and x = -3/2.

**Verifying the Relationship between Zeroes and Coefficients**

In a quadratic polynomial of the form ax^2 + bx + c, the relationship between the zeroes and the coefficients can be represented as follows:

- The sum of the zeroes is equal to -b/a.
- The product of the zeroes is equal to c/a.

In our case, the polynomial is x^2 + (1/6)x - 2. Comparing this with the general form ax^2 + bx + c, we can see that a = 1, b = 1/6, and c = -2.

- The sum of the zeroes is (-3/2) + (4/3) = 3/6 + 8/6 = 11/6. Comparing this with -b/a = -1/6, we can see that the sum of the zeroes is equal to -b/a.
- The product of the zeroes is (-3/2) * (4/3) = -12/6 = -2. Comparing this with c/a = -2, we can see that the product of the zeroes is equal to c/a.

Therefore, we have verified that the relationship between the zeroes and the coefficients holds true for the quadratic polynomial x^2 + (1/6)x - 2.