To find the quadratic polynomial given the product and sum of its zeroes, we can use Vieta's formulas.

Let the quadratic polynomial be of the form:

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

For a quadratic equation, the relationships between the coefficients and the zeroes (roots) $\alpha$ and $\beta$ are given by:

1. The sum of the zeroes $\alpha + \beta = -\frac{b}{a}$
2. The product of the zeroes $\alpha \beta = \frac{c}{a}$

From the problem, we know:

• The sum of the zeroes is $-2$, so: $\alpha + \beta = -2$

• The product of the zeroes is $\frac{1}{5}$, so: $\alpha \beta = \frac{1}{5}$

Now, using these relationships, we can express the quadratic equation as:

$f(x) = a(x^2 - (\alpha + \beta)x + \alpha \beta)$

Substituting the given values:

$f(x) = a(x^2 - (-2)x + \frac{1}{5})$

Simplifying:

$f(x) = a(x^2 + 2x + \frac{1}{5})$

To eliminate the fraction, multiply the entire equation by 5:

$5f(x) = 5a(x^2 + 2x + \frac{1}{5}) = a(5x^2 + 10x + 1)$

Thus, the quadratic polynomial becomes:

$f(x) = a(5x^2 + 10x + 1)$

Since the leading coefficient $a$ is arbitrary, we can set $a = 1$ for simplicity:

$f(x) = 5x^2 + 10x + 1$

Thus, the quadratic polynomial is:

$f(x) = 5x^2 + 10x + 1$

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Would you like further details on any part of this process, or do you have any other questions?

Here are some related questions:

1. How do you find the zeroes of a quadratic polynomial?
2. What is the significance of the discriminant in determining the number of real roots of a quadratic equation?
3. How can we modify the polynomial if the leading coefficient $a$ is different from 1?
4. Can we derive the sum and product of the zeroes from a given quadratic equation?
5. What would happen if the product of the zeroes was negative?

Tip: Always check the sign of the sum and product of zeroes to help determine the nature of the quadratic equation!