For the given sum and product of zeroes,find the quadratic polynomial. Also,find the zeroes of this polynomial by factorization:
Sum of zeroes = $-2 \sqrt{3}$,Product of zeroes = $-9$.

(N/A) Given that,sum of zeroes $(S)$ = $-2 \sqrt{3}$ and product of zeroes $(P)$ = $-9$.
The general form of a quadratic polynomial is $f(x) = x^{2} - Sx + P$.
Substituting the values,we get $f(x) = x^{2} - (-2 \sqrt{3})x + (-9) = x^{2} + 2 \sqrt{3}x - 9$.
To find the zeroes by factorization,we split the middle term $2 \sqrt{3}x$ into $3 \sqrt{3}x - \sqrt{3}x$:
$f(x) = x^{2} + 3 \sqrt{3}x - \sqrt{3}x - 9$
$f(x) = x(x + 3 \sqrt{3}) - \sqrt{3}(x + 3 \sqrt{3})$
$f(x) = (x + 3 \sqrt{3})(x - \sqrt{3})$
Setting $f(x) = 0$,we get $x + 3 \sqrt{3} = 0$ or $x - \sqrt{3} = 0$.
Therefore,the zeroes are $-3 \sqrt{3}$ and $\sqrt{3}$.