Let us solve the given polynomial $F(x) = x^3 - x^2 - 6x$ to find its zeros.

Step 1: Factor the polynomial

We can start by factoring out the greatest common factor (GCF): $F(x) = x(x^2 - x - 6).$

Step 2: Factor the quadratic

Now, factorize the quadratic $x^2 - x - 6$: $x^2 - x - 6 = (x - 3)(x + 2).$

Step 3: Write the fully factored form

The fully factored form of $F(x)$ is: $F(x) = x(x - 3)(x + 2).$

Step 4: Solve for the zeros

Set each factor equal to zero:

1. $x = 0$,
2. $x - 3 = 0$ → $x = 3$,
3. $x + 2 = 0$ → $x = -2$.

Thus, the zeros of the polynomial are: $x = -2, x = 0, \text{ and } x = 3.$

Final Answer:

The correct choice is: D. $x = -2, x = 0, \text{ and } x = 3$.

Do you have any questions or need further clarification?

Related Questions:

1. What does it mean for a value to be a "zero" of a polynomial function?
2. How do you factorize a quadratic equation quickly?
3. How do polynomial degrees relate to the number of zeros?
4. How can the graph of $F(x) = x^3 - x^2 - 6x$ confirm the zeros?
5. What are the steps to check your factored form is correct?

Tip:

Always double-check your factored expression by multiplying the factors back to ensure it equals the original polynomial!