Step 1: A quadratic polynomial with given sum and product of zeroes is
$$p(x)=x^{2}-(\text{sum})x+(\text{product})$$
Step 2: Substitute sum $=0$ and product $=-9$.
$$p(x)=x^{2}-(0)x+(-9)=x^{2}-9$$
Step 3: To find the zeroes, put $p(x)=0$.
$$x^{2}-9=0$$
Step 4: Factorise using $a^{2}-b^{2}=(a+b)(a-b)$.
$$(x+3)(x-3)=0$$
Step 5: Therefore
$$x=3 \quad\text{or}\quad x=-3$$
Step 6: Verification: sum $=3+(-3)=0$ ✓ and product $=3\times(-3)=-9$ ✓
Hence the polynomial is $x^{2}-9$ and its zeroes are $3$ and $-3$.