To solve this problem, let's break it down step-by-step.

We are given the polynomial:

$p x^2 - 4x - 2p$

We are told that the sum of the zeroes of the polynomial is the same as the product of the zeroes. To use this information, we'll need to recall the relations between the sum and product of the zeroes of a quadratic equation.

For a general quadratic equation of the form:

$ax^2 + bx + c$

The sum and product of its zeroes (roots), say $\alpha$ and $\beta$, are given by:

• Sum of the zeroes: $\alpha + \beta = -\frac{b}{a}$
• Product of the zeroes: $\alpha \beta = \frac{c}{a}$

Step 1: Identify coefficients

For the polynomial $p x^2 - 4x - 2p$, we can identify the coefficients:

• $a = p$
• $b = -4$
• $c = -2p$

Step 2: Apply the relations

Now, let's use the sum and product of the zeroes:

• The sum of the zeroes is:

$\text{Sum of the zeroes} = -\frac{b}{a} = -\frac{-4}{p} = \frac{4}{p}$

• The product of the zeroes is:

$\text{Product of the zeroes} = \frac{c}{a} = \frac{-2p}{p} = -2$

Step 3: Set up the equation

We are told that the sum of the zeroes is equal to the product of the zeroes:

$\frac{4}{p} = -2$

Step 4: Solve for $p$

Now, solve the equation for $p$:

$\frac{4}{p} = -2$

Multiply both sides by $p$:

$4 = -2p$

Now, divide both sides by -2:

$p = -2$

Final Answer:

The value of $p$ is $\boxed{-2}$.

Would you like a detailed breakdown of any part of this process? Here are a few related questions to explore further:

1. How would you solve for the zeroes of a quadratic equation once the value of $p$ is known?
2. What happens to the sum and product of the zeroes if the leading coefficient $a$ is 0?
3. How do you handle quadratics where the coefficients are more complex functions of $p$?
4. What is the relationship between the discriminant and the zeroes of a quadratic equation?
5. How does changing the value of $p$ affect the graph of the quadratic equation?

Tip: Always remember that the sum and product of the zeroes of a quadratic can be derived directly from the coefficients of the equation, which is a quick shortcut to solving many such problems.