A quadratic polynomial,

Solution:

Since $\alpha$ and $\beta$ are the zeros of the quadratic polynomials such that

$0=\alpha+\beta$

If one of zero is 3 then

$\alpha+\beta=0$

$3+\beta=0$

$\beta=0-3$

$\beta=-3$

Substituting $\beta=-3$ in $\alpha+\beta=0$ we get

$\alpha-3=0$

$\alpha=3$

Let S and P denote the sum and product of the zeros of the polynomial respectively then

$S=\alpha+\beta$

$S=0$

$P=\alpha \beta$

$P=3 \times-3$

$P=-9$

Hence, the required polynomials is

$=\left(x^{2}-S x+P\right)$

$=\left(x^{2}-0 x-9\right)$

$=x^{2}-9$

Hence, the correct choice is (a)