We are asked to find the polynomial $P(x) = x^3 + ax^2 + bx + c$ such that it satisfies the following conditions:

1. $x = -1$ is a local maximum.
2. $x = 5$ is a local minimum.
3. $P(0) = 0$.

Steps:

1. Derivative of $P(x)$: To find local extrema, we take the first derivative of $P(x)$: $P'(x) = 3x^2 + 2ax + b$ Since $x = -1$ is a local maximum and $x = 5$ is a local minimum, we set $P'(-1) = 0$ and $P'(5) = 0$.

2. Setting up equations from the derivative:

   • For $x = -1$ (local maximum): $P'(-1) = 3(-1)^2 + 2a(-1) + b = 0 \implies 3 - 2a + b = 0 \implies b = 2a - 3$
   • For $x = 5$ (local minimum): $P'(5) = 3(5)^2 + 2a(5) + b = 0 \implies 75 + 10a + b = 0 \implies b = -10a - 75$

   Now, we have two expressions for $b$: $2a - 3 = -10a - 75$ Solving for $a$: $2a + 10a = -75 + 3 \implies 12a = -72 \implies a = -6$ Substituting $a = -6$ into one of the expressions for $b$: $b = 2(-6) - 3 = -12 - 3 = -15$

3. Condition $P(0) = 0$: Using the condition $P(0) = 0$ in the original equation: $P(0) = 0^3 + a(0)^2 + b(0) + c = c$ Therefore, $c = 0$.

Final Polynomial:

Substituting $a = -6$, $b = -15$, and $c = 0$ into the polynomial: $P(x) = x^3 - 6x^2 - 15x$

Thus, the polynomial is: $P(x) = x^3 - 6x^2 - 15x$

Would you like to see more details on how I solved this? Here are some related questions:

1. How do we determine whether a point is a local maximum or minimum using derivatives?
2. What is the significance of the second derivative in verifying the nature of critical points?
3. How would the solution change if the polynomial had a higher degree?
4. Can you explain why $P'(x) = 0$ at critical points leads to local maxima or minima?
5. How can you graphically represent the behavior of this polynomial?

Tip: When solving polynomial problems with local maxima and minima, always remember that these points occur where the derivative equals zero!