find zeroes of the given quadratic polynomials and verify the relation between zeroes and coefficients :
`a^(2)x^(2)-11a^(2)x+30a^(2)-a-1,a ne 0`

Explanation

To find the zeroes of the quadratic polynomial \( a^2 x^2 - 11a^2 x + (30a^2 - a - 1) \) and verify the relation between the zeroes and coefficients, we will follow these steps:

### Step 1: Identify the coefficients
The given polynomial is in the form of \( ax^2 + bx + c \). Here:
- \( p = a^2 \)
- \( q = -11a^2 \)
- \( r = 30a^2 - a - 1 \)

### Step 2: Calculate the discriminant
The discriminant \( D \) of a quadratic equation is given by the formula:
\[ D = q^2 - 4pr \]
Substituting the values of \( q \), \( p \), and \( r \):
\[ D = (-11a^2)^2 - 4(a^2)(30a^2 - a - 1) \]
Calculating \( D \):
\[ D = 121a^4 - 4a^2(30a^2 - a - 1) \]
Expanding the second term:
\[ D = 121a^4 - (120a^4 - 4a^3 - 4a^2) \]
Combining like terms:
\[ D = 121a^4 - 120a^4 + 4a^3 + 4a^2 = a^4 + 4a^3 + 4a^2 \]
Factoring out \( a^2 \):
\[ D = a^2(a^2 + 4a + 4) = a^2(a + 2)^2 \]

### Step 3: Find the roots using the quadratic formula
The roots (zeroes) of the quadratic polynomial can be found using the quadratic formula:
\[ x = \frac{-q \pm \sqrt{D}}{2p} \]
Substituting the values of \( q \), \( D \), and \( p \):
\[ x = \frac{-(-11a^2) \pm \sqrt{a^2(a + 2)^2}}{2a^2} \]
This simplifies to:
\[ x = \frac{11a^2 \pm a(a + 2)}{2a^2} \]
This gives us two cases for \( x \):
1. \( x_1 = \frac{11a^2 + a(a + 2)}{2a^2} \)
2. \( x_2 = \frac{11a^2 - a(a + 2)}{2a^2} \)

Calculating \( x_1 \):
\[ x_1 = \frac{11a^2 + a^2 + 2a}{2a^2} = \frac{12a^2 + 2a}{2a^2} = \frac{6a + 1}{a} \]

Calculating \( x_2 \):
\[ x_2 = \frac{11a^2 - a^2 - 2a}{2a^2} = \frac{10a^2 - 2a}{2a^2} = \frac{5a - 1}{a} \]

Thus, the zeroes are:
\[ x_1 = \frac{6a + 1}{a}, \quad x_2 = \frac{5a - 1}{a} \]

### Step 4: Verify the relation between zeroes and coefficients
The sum of the roots \( x_1 + x_2 \) should equal \( -\frac{q}{p} \):
\[ x_1 + x_2 = \frac{6a + 1}{a} + \frac{5a - 1}{a} = \frac{(6a + 1) + (5a - 1)}{a} = \frac{11a}{a} = 11 \]
And \( -\frac{q}{p} = -\frac{-11a^2}{a^2} = 11 \). Thus, the sum of the roots is verified.

Now, the product of the roots \( x_1 \cdot x_2 \) should equal \( \frac{r}{p} \):
\[ x_1 \cdot x_2 = \left(\frac{6a + 1}{a}\right) \cdot \left(\frac{5a - 1}{a}\right) = \frac{(6a + 1)(5a - 1)}{a^2} \]
Calculating the product:
\[ = \frac{30a^2 - 6a + 5a - 1}{a^2} = \frac{30a^2 - a - 1}{a^2} \]
And \( \frac{r}{p} = \frac{30a^2 - a - 1}{a^2} \). Thus, the product of the roots is verified.

### Final Result
The zeroes of the polynomial are:
\[ x_1 = \frac{6a + 1}{a}, \quad x_2 = \frac{5a - 1}{a} \]
And the relations between the zeroes and coefficients are verified.