x^2(x+1)^3/(x-7)(x+3)^2(-x^2-1) is less than or equal to 0

We need to solve the inequality:

\[
\frac{x^2 (x+1)^3}{(x-7)(x+3)^2 (-x^2 - 1)} \leq 0
\]

### Step 1: Simplify the expression
Let's first simplify each component of the expression. We can note the following:

- The term \( -x^2 - 1 \) can be written as \( -(x^2 + 1) \). Since \( x^2 + 1 \) is always positive for all real \(x\), the entire term is always negative, i.e. \( -(x^2 + 1) < 0 \).

The inequality now becomes:

\[
\frac{x^2 (x+1)^3}{(x-7)(x+3)^2} \geq 0
\]

### Step 2: Analyze the critical points
Next, we find the points where the expression is equal to zero or undefined. These points occur when the numerator is zero or the denominator is undefined.

#### Critical points:
- \(x^2 = 0\) gives \(x = 0\).
- \((x+1)^3 = 0\) gives \(x = -1\).
- \(x-7 = 0\) gives \(x = 7\).
- \((x+3)^2 = 0\) gives \(x = -3\).

These are the key values: \(x = -3, -1, 0, 7\).

### Step 3: Test the sign of the expression in intervals
We now break the real number line into intervals determined by these critical points: \( (-\infty, -3), (-3, -1), (-1, 0), (0, 7), (7, \infty) \). In each interval, we'll check the sign of the expression.

#### Intervals:
1. **Interval \((- \infty, -3)\)**:
   - Pick \(x = -4\). In this case, \(x^2 > 0\), \((x+1)^3 < 0\), \((x-7) < 0\), and \((x+3)^2 > 0\).
   - The sign of the expression: \(\frac{(positive)(negative)}{(negative)(positive)} = positive\).

2. **Interval \((-3, -1)\)**:
   - Pick \(x = -2\). In this case, \(x^2 > 0\), \((x+1)^3 < 0\), \((x-7) < 0\), and \((x+3)^2 > 0\).
   - The sign of the expression: \(\frac{(positive)(negative)}{(negative)(positive)} = positive\).

3. **Interval \((-1, 0)\)**:
   - Pick \(x = -0.5\). In this case, \(x^2 > 0\), \((x+1)^3 > 0\), \((x-7) < 0\), and \((x+3)^2 > 0\).
   - The sign of the expression: \(\frac{(positive)(positive)}{(negative)(positive)} = negative\).

4. **Interval \((0, 7)\)**:
   - Pick \(x = 1\). In this case, \(x^2 > 0\), \((x+1)^3 > 0\), \((x-7) < 0\), and \((x+3)^2 > 0\).
   - The sign of the expression: \(\frac{(positive)(positive)}{(negative)(positive)} = negative\).

5. **Interval \((7, \infty)\)**:
   - Pick \(x = 8\). In this case, \(x^2 > 0\), \((x+1)^3 > 0\), \((x-7) > 0\), and \((x+3)^2 > 0\).
   - The sign of the expression: \(\frac{(positive)(positive)}{(positive)(positive)} = positive\).

### Step 4: Evaluate at critical points
- At \(x = -3\), the denominator becomes zero, so the expression is undefined.
- At \(x = -1\), the numerator is zero, so the expression equals 0.
- At \(x = 0\), the numerator is zero, so the expression equals 0.
- At \(x = 7\), the denominator becomes zero, so the expression is undefined.

### Step 5: Conclusion
The solution to the inequality is where the expression is less than or equal to zero. This occurs in the intervals where the expression is negative or zero:

- The inequality is satisfied on \((-1, 0]\) and \([7, \infty)\).

Thus, the solution is:

\[
x \in [-1, 0] \cup (7, \infty)
\]

Would you like further details or explanations on any step?

Here are five related questions you could explore:
1. How do you solve inequalities involving rational functions?
2. Why is the expression undefined at certain points?
3. What are the key differences between strict and non-strict inequalities?
4. How do the multiplicities of roots affect the sign of an expression?
5. How would the solution change if we had a strict inequality \(<\) instead of \(\leq\)?

**Tip:** Always test points between critical values to determine the sign of the expression in different intervals!

Learn how to solve the rational inequality x^2(x+1)^3/(x-7)(x+3)^2(-x^2-1) ≤ 0 using critical points and sign analysis. This solution guides you through simplifying the expression, identifying critical points, and testing intervals. Suitable for high school students in Grades 10-12.

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