The given profit function is:

$P(x) = -2x^2 + 60x - 250$

We are asked to determine two things:

1. The factors that can be used to find the zeroes of the function (i.e., the points where $P(x) = 0$).
2. The number of units the company should make and sell to earn a profit.

Step 1: Finding the factors

To find the factors, we need to solve for $x$ when $P(x) = 0$.

$-2x^2 + 60x - 250 = 0$

First, divide the whole equation by $-2$ to simplify:

$x^2 - 30x + 125 = 0$

Next, we solve this quadratic equation using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

For the equation $x^2 - 30x + 125 = 0$, we have:

• $a = 1$
• $b = -30$
• $c = 125$

Substitute these values into the quadratic formula:

$x = \frac{-(-30) \pm \sqrt{(-30)^2 - 4(1)(125)}}{2(1)}$

$x = \frac{30 \pm \sqrt{900 - 500}}{2}$

$x = \frac{30 \pm \sqrt{400}}{2}$

$x = \frac{30 \pm 20}{2}$

Thus, the two solutions are:

$x = \frac{30 + 20}{2} = 25 \quad \text{and} \quad x = \frac{30 - 20}{2} = 5$

So the factors are $(x - 25)$ and $(x - 5)$.

Step 2: Finding the interval for profit

For the company to make a profit, it needs to produce units between the zeroes of the function. The zeroes are $x = 5$ and $x = 25$, so the company should make and sell between 5 and 25 units to earn a profit.

Final Answers:

• Factors: $(x - 25)$ and $(x - 5)$
• Units to make and sell: Between 5 and 25 units

Would you like further details or have any questions?

Here are 5 related questions to expand on this:

1. How do you determine whether a quadratic function opens upward or downward?
2. What is the significance of the vertex in the profit function?
3. Can you explain how to complete the square for this quadratic function?
4. How does solving the quadratic equation help in business decision-making?
5. What is the relationship between the roots of the quadratic function and the break-even points?

Tip: The discriminant $\Delta = b^2 - 4ac$ of a quadratic equation tells you the nature of the roots (real or complex).