To find a quadratic polynomial whose zeroes are $-3$ and $4$, we can use the fact that a quadratic polynomial can be written in terms of its roots as:

$f(x) = a(x - r_1)(x - r_2)$

where $r_1$ and $r_2$ are the roots (zeroes) of the polynomial, and $a$ is a constant coefficient.

In this case, the roots are $r_1 = -3$ and $r_2 = 4$. So the polynomial is:

$f(x) = a(x - (-3))(x - 4) = a(x + 3)(x - 4)$

Now, expand the expression:

$f(x) = a(x^2 - 4x + 3x - 12) = a(x^2 - x - 12)$

For simplicity, we can take $a = 1$ (unless otherwise specified). So the quadratic polynomial is:

$f(x) = x^2 - x - 12$

This is the required quadratic polynomial whose zeroes are $-3$ and $4$.

Would you like further details or have any questions about this?

Here are some related questions for you to explore:

1. How do you find the sum and product of the roots from the quadratic formula?
2. What would happen if one of the zeroes is complex? How does the polynomial change?
3. How can you determine the roots of a quadratic polynomial from its general form?
4. How do you factor a quadratic expression like $x^2 - x - 12$?
5. Can you use the quadratic formula to find the roots of any polynomial, even non-quadratic ones?

Tip: Always remember that the sum of the roots of a quadratic polynomial is equal to $-b/a$ and the product of the roots is equal to $c/a$ (from the standard quadratic form $ax^2 + bx + c$).