Find zeroes of the given quadratic polynomials and verify the relation between zeroes and coefficients :
`6x^(2)-x-2`

Explanation

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

### Step 1: Identify the coefficients
The given polynomial is \(6x^2 - x - 2\). We can identify the coefficients as follows:
- \(a = 6\) (coefficient of \(x^2\))
- \(b = -1\) (coefficient of \(x\))
- \(c = -2\) (constant term)

### Step 2: Set the polynomial equal to zero
To find the zeroes, we set the polynomial equal to zero:
\[ 6x^2 - x - 2 = 0 \]

### Step 3: Factor the quadratic equation
We will factor the quadratic equation. We need to find two numbers that multiply to \(a \cdot c = 6 \cdot (-2) = -12\) and add to \(b = -1\). The numbers that satisfy this are \(3\) and \(-4\).

Now we can rewrite the middle term:
\[ 6x^2 + 3x - 4x - 2 = 0 \]

Next, we group the terms:
\[ (6x^2 + 3x) + (-4x - 2) = 0 \]

Factor by grouping:
\[ 3x(2x + 1) - 2(2x + 1) = 0 \]

Now we can factor out the common factor \((2x + 1)\):
\[ (2x + 1)(3x - 2) = 0 \]

### Step 4: Solve for the zeroes
We set each factor equal to zero:
1. \(2x + 1 = 0\)
\[ 2x = -1 \implies x = -\frac{1}{2} \]
2. \(3x - 2 = 0\)
\[ 3x = 2 \implies x = \frac{2}{3} \]

Thus, the zeroes of the polynomial are:
\[ x_1 = -\frac{1}{2}, \quad x_2 = \frac{2}{3} \]

### Step 5: Verify the relation between zeroes and coefficients
The sum of the zeroes (\(x_1 + x_2\)) is given by:
\[ x_1 + x_2 = -\frac{1}{2} + \frac{2}{3} \]
To add these fractions, we find a common denominator (which is 6):
\[ -\frac{1}{2} = -\frac{3}{6}, \quad \frac{2}{3} = \frac{4}{6} \]
So,
\[ x_1 + x_2 = -\frac{3}{6} + \frac{4}{6} = \frac{1}{6} \]
According to the relation, the sum of the zeroes is given by \(-\frac{b}{a}\):
\[ -\frac{b}{a} = -\frac{-1}{6} = \frac{1}{6} \]

Now, for the product of the zeroes (\(x_1 \cdot x_2\)):
\[ x_1 \cdot x_2 = -\frac{1}{2} \cdot \frac{2}{3} = -\frac{1}{3} \]
According to the relation, the product of the zeroes is given by \(\frac{c}{a}\):
\[ \frac{c}{a} = \frac{-2}{6} = -\frac{1}{3} \]

### Conclusion
Both the sum and product of the zeroes match the relations derived from the coefficients. Therefore, we have verified the relation between the zeroes and coefficients.