For each of the following find a quadratic polynomial whose sum and product respectively of the zeroes are as given .Also find the zeroes of these polynomials by factorisation.
`-2sqrt(3),-9`

Answer

To find a quadratic polynomial whose zeros have a given sum and product, we can use the standard form of a quadratic polynomial, which is:

\[ P(x) = x^2 - (sum \ of \ zeros) \cdot x + (product \ of \ zeros) \]

### Step 1: Identify the sum and product of the zeros
We are given:
- Sum of the zeros \( S = -2\sqrt{3} \)
- Product of the zeros \( P = -9 \)

### Step 2: Write the quadratic polynomial
Using the formula mentioned above, we can substitute the values of the sum and product into the polynomial:

\[ P(x) = x^2 - (-2\sqrt{3}) \cdot x + (-9) \]

This simplifies to:

\[ P(x) = x^2 + 2\sqrt{3}x - 9 \]

### Step 3: Factor the polynomial
Now, we need to factor the polynomial \( P(x) = x^2 + 2\sqrt{3}x - 9 \).

To factor, we look for two numbers that multiply to \( -9 \) (the product) and add to \( 2\sqrt{3} \) (the sum).

Let's rewrite the middle term \( 2\sqrt{3}x \) as \( 3\sqrt{3}x - \sqrt{3}x \):

\[ P(x) = x^2 + 3\sqrt{3}x - \sqrt{3}x - 9 \]

Now, we can group the terms:

\[ P(x) = (x^2 + 3\sqrt{3}x) + (-\sqrt{3}x - 9) \]

Factoring out common terms from each group:

\[ P(x) = x(x + 3\sqrt{3}) - \sqrt{3}(x + 3) \]

Now we can factor by grouping:

\[ P(x) = (x + 3\sqrt{3})(x - \sqrt{3}) \]

### Step 4: Find the zeros of the polynomial
To find the zeros, we set each factor equal to zero:

1. \( x + 3\sqrt{3} = 0 \)
- \( x = -3\sqrt{3} \)

2. \( x - \sqrt{3} = 0 \)
- \( x = \sqrt{3} \)

Thus, the zeros of the polynomial are:

- \( x = -3\sqrt{3} \)
- \( x = \sqrt{3} \)

### Summary
The quadratic polynomial is:

\[ P(x) = x^2 + 2\sqrt{3}x - 9 \]

And the zeros of this polynomial are:

- \( x = -3\sqrt{3} \)
- \( x = \sqrt{3} \)