The general form of a quadratic polynomial is:

$P(x) = ax^2 + bx + c$

According to Vieta's formulas:

• The sum of the zeroes of the quadratic polynomial is given by $-\frac{b}{a}$.
• The product of the zeroes is $\frac{c}{a}$.

Given that:

1. The sum of the zeroes is 0.
2. One of the zeroes is 3.

Let's call the other zero $r$. Based on the first condition:

$3 + r = 0$

So, $r = -3$.

Thus, the two zeroes are 3 and $-3$.

Now, the quadratic polynomial can be written as:

$P(x) = a(x - 3)(x + 3)$

Simplifying this:

$P(x) = a(x^2 - 9)$

Since no specific value for $a$ is given, the polynomial could be $P(x) = a(x^2 - 9)$, where $a$ is any constant.

If we assume $a = 1$, the polynomial becomes:

$P(x) = x^2 - 9$

Would you like more details or have any questions?

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Here are some related questions you might find helpful:

1. What are Vieta's formulas for a quadratic equation?
2. How can you find the product of the zeroes of a quadratic polynomial?
3. What does the coefficient $a$ represent in a quadratic equation?
4. How can you determine the discriminant of a quadratic equation?
5. Can a quadratic polynomial have zero as a root?

Tip: For any quadratic equation, the sum and product of the roots can be found using the coefficients of $x$ and the constant term!