Find zeroes of the polynormial `6x^(2)-3-7x` and verify the relation between zeroes and coefficients.

Explanation

To find the zeroes of the polynomial \(6x^2 - 3 - 7x\) and verify the relation between the zeroes and coefficients, we can follow these steps:

### Step 1: Arrange the polynomial
First, we rewrite the polynomial in standard form:
\[ 6x^2 - 7x - 3 \]

### Step 2: Identify coefficients
From the polynomial \(6x^2 - 7x - 3\), we identify the coefficients:
- \(a = 6\)
- \(b = -7\)
- \(c = -3\)

### Step 3: Factor the polynomial
To find the zeroes, we can factor the polynomial. We will use the method of splitting the middle term. We need two numbers that multiply to \(a \cdot c = 6 \cdot (-3) = -18\) and add to \(b = -7\).

The two numbers that satisfy these conditions are \(-9\) and \(2\).

Now, we can rewrite the polynomial:
\[ 6x^2 - 9x + 2x - 3 \]

### Step 4: Group the terms
Next, we group the terms:
\[ (6x^2 - 9x) + (2x - 3) \]

### Step 5: Factor by grouping
Now we factor out the common factors in each group:
\[ 3x(2x - 3) + 1(2x - 3) \]

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

### Step 6: Set the factors to zero
To find the zeroes, we set each factor equal to zero:
1. \(2x - 3 = 0\)
2. \(3x + 1 = 0\)

### Step 7: Solve for \(x\)
Solving these equations gives:
1. \(2x - 3 = 0 \implies 2x = 3 \implies x = \frac{3}{2}\)
2. \(3x + 1 = 0 \implies 3x = -1 \implies x = -\frac{1}{3}\)

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

### Step 8: Verify the relation between zeroes and coefficients
We need to verify two relations:
1. The sum of the zeroes should equal \(-\frac{b}{a}\).
2. The product of the zeroes should equal \(\frac{c}{a}\).

#### Sum of the zeroes:
\[ \text{Sum} = \frac{3}{2} + \left(-\frac{1}{3}\right) = \frac{3}{2} - \frac{1}{3} \]
Finding a common denominator (which is 6):
\[ \text{Sum} = \frac{9}{6} - \frac{2}{6} = \frac{7}{6} \]
Now, calculate \(-\frac{b}{a}\):
\[ -\frac{-7}{6} = \frac{7}{6} \]
Thus, the sum of the zeroes is verified.

#### Product of the zeroes:
\[ \text{Product} = \frac{3}{2} \cdot \left(-\frac{1}{3}\right) = -\frac{1}{2} \]
Now, calculate \(\frac{c}{a}\):
\[ \frac{-3}{6} = -\frac{1}{2} \]
Thus, the product of the zeroes is verified.

### Final Answer
The zeroes of the polynomial \(6x^2 - 3 - 7x\) are \(x = \frac{3}{2}\) and \(x = -\frac{1}{3}\). The relations between the zeroes and coefficients are verified.