If the product of zeros of `ax^(2)-6x-6` is `4`, find the value of `a`. Hence find the sum of its zeros.

Answer

To solve the given problem, we need to find the value of \(a\) and then use it to find the sum of the zeros of the quadratic polynomial \(ax^2 - 6x - 6\).

### Step-by-Step Solution

1. **Identify the given polynomial and the product of its zeros:**

The given polynomial is:
\[ ax^2 - 6x - 6 \]
The product of the zeros is given as \(4\).

2. **Recall the standard form of a quadratic polynomial and the formulas for the sum and product of its zeros:**

A general quadratic polynomial is:
\[ ax^2 + bx + c \]
For this polynomial:
- The sum of the zeros (\(\alpha + \beta\)) is given by \(-\frac{b}{a}\).
- The product of the zeros (\(\alpha \beta\)) is given by \(\frac{c}{a}\).

3. **Compare the given polynomial with the standard form:**

For the polynomial \(ax^2 - 6x - 6\):
- \(a = a\)
- \(b = -6\)
- \(c = -6\)

4. **Use the product of the zeros formula to find \(a\):**

Given that the product of the zeros is \(4\), we use the formula:
\[ \alpha \beta = \frac{c}{a} \]
Substituting the values:
\[ 4 = \frac{-6}{a} \]
Solving for \(a\):
\[ 4a = -6 \]
\[ a = \frac{-6}{4} = -\frac{3}{2} \]

5. **Find the sum of the zeros using the value of \(a\):**

The sum of the zeros is given by:
\[ \alpha + \beta = -\frac{b}{a} \]
Substituting the values:
\[ \alpha + \beta = -\left(\frac{-6}{-\frac{3}{2}}\right) \]
Simplifying:
\[ \alpha + \beta = \frac{6}{\frac{3}{2}} = 6 \times \frac{2}{3} = 4 \]
Since we have a negative sign in the denominator:
\[ \alpha + \beta = -4 \]

### Final Answer
- The value of \(a\) is \(-\frac{3}{2}\).
- The sum of the zeros is \(-4\).