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The inequality to solve is:

\[
\frac{(2x + 1)^2}{(x - 10)(x + 10)} \leq 0
\]

### Step 1: Identify the critical points
The critical points occur when the numerator or denominator is equal to zero:
1. **Numerator:** \((2x + 1)^2 = 0 \Rightarrow 2x + 1 = 0 \Rightarrow x = -\frac{1}{2}\)
2. **Denominator:** 
   \[
   (x - 10)(x + 10) = 0 \Rightarrow x = 10 \text{ or } x = -10
   \]

The critical points are: \(x = -\frac{1}{2}, x = -10, x = 10\).

### Step 2: Analyze intervals
The critical points divide the real line into four intervals:
- \( (-\infty, -10) \)
- \( (-10, -\frac{1}{2}) \)
- \( \left(-\frac{1}{2}, 10\right) \)
- \( (10, \infty) \)

We will check the sign of the expression in each interval.

### Step 3: Test each interval
The expression is:
\[
\frac{(2x + 1)^2}{(x - 10)(x + 10)}
\]
Since \((2x + 1)^2\) is always non-negative (it’s a square), the sign of the overall expression depends on the denominator \((x - 10)(x + 10)\).

1. **For \(x \in (-\infty, -10)\):**
   Choose \(x = -20\):
   \((2x + 1)^2 = (2(-20) + 1)^2 = (-39)^2 > 0\), and \((x - 10)(x + 10) = (-20 - 10)(-20 + 10) = (-30)(-10) > 0\).
   Hence, the expression is positive.

2. **For \(x \in (-10, -\frac{1}{2})\):**
   Choose \(x = -1\):
   \((2x + 1)^2 = (2(-1) + 1)^2 = (-1)^2 = 1 > 0\), and \((x - 10)(x + 10) = (-1 - 10)(-1 + 10) = (-11)(9) < 0\).
   Hence, the expression is negative.

3. **For \(x \in \left(-\frac{1}{2}, 10\right)\):**
   Choose \(x = 0\):
   \((2x + 1)^2 = (2(0) + 1)^2 = 1^2 = 1 > 0\), and \((x - 10)(x + 10) = (0 - 10)(0 + 10) = (-10)(10) < 0\).
   Hence, the expression is negative.

4. **For \(x \in (10, \infty)\):**
   Choose \(x = 20\):
   \((2x + 1)^2 = (2(20) + 1)^2 = (41)^2 > 0\), and \((x - 10)(x + 10) = (20 - 10)(20 + 10) = (10)(30) > 0\).
   Hence, the expression is positive.

### Step 4: Evaluate at critical points
- At \(x = -10\): The denominator is zero, so it is undefined.
- At \(x = 10\): The denominator is also zero, so it is undefined.
- At \(x = -\frac{1}{2}\): The numerator is zero, so the entire expression is \(0\).

### Step 5: Solution in interval notation
The solution to \(\frac{(2x + 1)^2}{(x - 10)(x + 10)} \leq 0\) includes the intervals where the expression is negative or zero:
\[
(-10, -\frac{1}{2}] \cup \left(-\frac{1}{2}, 10\right)
\]

The solution set is:
\[
(-10, -\frac{1}{2}] \cup \left(-\frac{1}{2}, 10\right)
\]

Would you like further details or have any questions?

### 5 Related Questions
1. How would the solution change if the inequality was \(\geq 0\)?
2. What happens if you change the numerator to \((2x + 1)^3\)?
3. How do you solve similar rational inequalities with a different sign, such as \(>\) or \(<\)?
4. Can you explain how interval notation works in more detail?
5. What role does the multiplicity of roots play in determining the sign of a rational expression?

### Tip
When solving rational inequalities, always identify where the numerator and denominator are zero, as these points are critical for defining the intervals.

Solve the inequality (2x + 1)^2 / [(x - 10)(x + 10)] <= 0. State the solution set using interval notation.

Solve the rational inequality (2x + 1)^2 / [(x - 10)(x + 10)] <= 0 using interval notation. This problem focuses on critical points, sign analysis, and rational expressions. Suitable for students in Grades 9-11.

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