The given polynomial inequality is:

$-x^3 - x^2 + 30x > 0$

Let's solve it step by step.

1. Factor the polynomial

We begin by factoring the left-hand side:

$-x^3 - x^2 + 30x = -x(x^2 + x - 30)$

Now, let's factor the quadratic $x^2 + x - 30$. To do this, we need to find two numbers whose product is $-30$ and whose sum is $1$. These numbers are $6$ and $-5$. Thus, we can factor the quadratic as:

$x^2 + x - 30 = (x - 5)(x + 6)$

Substituting this back into the original equation:

$-x(x - 5)(x + 6) > 0$

2. Determine the critical points

The critical points are the values of $x$ where the expression equals zero. These occur when any factor of the inequality equals zero. So, we solve:

$-x = 0 \quad \text{or} \quad x - 5 = 0 \quad \text{or} \quad x + 6 = 0$

This gives the critical points:

$x = 0, \quad x = 5, \quad x = -6$

3. Analyze the sign of the expression in each interval

The critical points divide the real line into four intervals:

• $(-\infty, -6)$
• $(-6, 0)$
• $(0, 5)$
• $(5, \infty)$

We now test the sign of $-x(x - 5)(x + 6)$ in each interval by picking test points:

• For $x = -7$ (in $(-∞, -6)$): $-(-7)((-7) - 5)((-7) + 6) = -(-7)(-12)(-1) = -(-7)(12)(1) = 84 > 0$
• For $x = -1$ (in $(-6, 0)$): $-(-1)((-1) - 5)((-1) + 6) = -(-1)(-6)(5) = -(1)(6)(5) = -30 < 0$
• For $x = 2$ (in $(0, 5)$): $-(2)((2) - 5)((2) + 6) = -(2)(-3)(8) = -(2)(-24) = 48 > 0$
• For $x = 6$ (in $(5, ∞)$): $-(6)((6) - 5)((6) + 6) = -(6)(1)(12) = -72 < 0$

4. Determine the solution

We are looking for where the expression is positive. Based on our test points:

• Positive on $(-∞, -6)$
• Negative on $(-6, 0)$
• Positive on $(0, 5)$
• Negative on $(5, ∞)$

5. Final solution

Thus, the solution to the inequality $-x(x - 5)(x + 6) > 0$ is:

$x \in (-\infty, -6) \cup (0, 5)$

Would you like a more detailed explanation on any step or have any questions?

Here are five related questions:

1. How do you factor quadratic expressions more generally?
2. How can we solve polynomial inequalities with higher degrees?
3. What is the graphical interpretation of solving polynomial inequalities?
4. Why do we test intervals around critical points in inequalities?
5. How do changes in sign affect the solution to polynomial inequalities?

Tip: When solving inequalities, always test the sign in each interval between critical points to ensure accuracy in your solution!