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 zeroes and coefficients, we can follow these steps:

### Step 1: Rewrite the Polynomial
First, we rewrite the polynomial in standard form:
\[ p(x) = 6x^2 - 7x - 3 \]

### Step 2: Factor the Polynomial
Next, we need to factor the polynomial \(6x^2 - 7x - 3\). We look for two numbers that multiply to \(6 \times (-3) = -18\) and add to \(-7\). The numbers \(-9\) and \(2\) work because:
\[ -9 \times 2 = -18 \quad \text{and} \quad -9 + 2 = -7 \]

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

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

Factoring out the common terms:
\[ 3x(2x - 3) + 1(2x - 3) \]

Now we can factor by grouping:
\[ (3x + 1)(2x - 3) \]

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

Solving these equations gives:
1. \(3x + 1 = 0 \Rightarrow 3x = -1 \Rightarrow x = -\frac{1}{3}\)
2. \(2x - 3 = 0 \Rightarrow 2x = 3 \Rightarrow x = \frac{3}{2}\)

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

### Step 4: Verify the Relation Between Zeroes and Coefficients
Now, we will verify the relations between the zeroes and the coefficients.

#### Sum of the Zeroes
The sum of the zeroes \(x_1 + x_2\) should equal \(-\frac{b}{a}\):
\[ x_1 + x_2 = \frac{3}{2} - \frac{1}{3} \]
To add these fractions, we find a common denominator (which is 6):
\[ \frac{3}{2} = \frac{9}{6}, \quad -\frac{1}{3} = -\frac{2}{6} \]
Thus,
\[ x_1 + x_2 = \frac{9}{6} - \frac{2}{6} = \frac{7}{6} \]

Now, using the coefficients \(a = 6\) and \(b = -7\):
\[ -\frac{b}{a} = -\frac{-7}{6} = \frac{7}{6} \]

This verifies that:
\[ x_1 + x_2 = -\frac{b}{a} \]

#### Product of the Zeroes
The product of the zeroes \(x_1 \cdot x_2\) should equal \(\frac{c}{a}\):
\[ x_1 \cdot x_2 = \frac{3}{2} \cdot -\frac{1}{3} = -\frac{3}{6} = -\frac{1}{2} \]

Using the coefficients \(c = -3\) and \(a = 6\):
\[ \frac{c}{a} = \frac{-3}{6} = -\frac{1}{2} \]

This verifies that:
\[ x_1 \cdot x_2 = \frac{c}{a} \]

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