Find zeroes of the given quadratic polynomials and verify the relation between zeroes and coefficients :
`6x^(2)+5ax-6a^(2)=0`

Explanation

To find the zeroes of the quadratic polynomial \(6x^2 + 5ax - 6a^2 = 0\) and verify the relation between the zeroes and coefficients, we can follow these steps:

### Step 1: Identify the coefficients
The given quadratic polynomial is in the form \(ax^2 + bx + c = 0\). Here:
- \(a = 6\)
- \(b = 5a\)
- \(c = -6a^2\)

### Step 2: Factor the polynomial
We can rewrite the polynomial to factor it. We need to express \(5ax\) in a way that allows us to factor by grouping:
\[ 6x^2 + 9ax - 4ax - 6a^2 = 0 \]
Now, we can group the terms:
\[ (6x^2 + 9ax) + (-4ax - 6a^2) = 0 \]
Factoring out the common terms:
\[ 3x(2x + 3a) - 2a(2x + 3a) = 0 \]
Now, we can factor out \((2x + 3a)\):
\[ (2x + 3a)(3x - 2a) = 0 \]

### Step 3: Solve for zeroes
Setting each factor to zero gives us the zeroes:
1. \(2x + 3a = 0 \Rightarrow x = -\frac{3a}{2}\)
2. \(3x - 2a = 0 \Rightarrow x = \frac{2a}{3}\)

Thus, the zeroes of the polynomial are:
\[ x_1 = -\frac{3a}{2}, \quad x_2 = \frac{2a}{3} \]

### Step 4: Calculate the sum of the zeroes
The sum of the zeroes \(x_1 + x_2\) is:
\[ -\frac{3a}{2} + \frac{2a}{3} \]
To add these fractions, we need a common denominator. The least common multiple of 2 and 3 is 6:
\[ -\frac{3a}{2} = -\frac{9a}{6}, \quad \frac{2a}{3} = \frac{4a}{6} \]
Thus,
\[ x_1 + x_2 = -\frac{9a}{6} + \frac{4a}{6} = -\frac{5a}{6} \]

### Step 5: Calculate the product of the zeroes
The product of the zeroes \(x_1 \cdot x_2\) is:
\[ -\frac{3a}{2} \cdot \frac{2a}{3} = -\frac{3a \cdot 2a}{2 \cdot 3} = -a^2 \]

### Step 6: Verify the relations
According to the relations for a quadratic polynomial:
- The sum of the zeroes is given by \(-\frac{b}{a}\)
- The product of the zeroes is given by \(\frac{c}{a}\)

For our polynomial:
- Sum of zeroes: \(-\frac{b}{a} = -\frac{5a}{6}\)
- Product of zeroes: \(\frac{c}{a} = \frac{-6a^2}{6} = -a^2\)

### Conclusion
We have found the zeroes:
\[ x_1 = -\frac{3a}{2}, \quad x_2 = \frac{2a}{3} \]
And verified the relations:
- Sum of zeroes: \(-\frac{5a}{6}\) matches \(-\frac{b}{a}\)
- Product of zeroes: \(-a^2\) matches \(\frac{c}{a}\)