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The given inequality is:

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

To solve this inequality, we need to analyze the sign of the expression \(\frac{(x + 2)(x - 7)}{(x + 3)^2}\) by considering the critical points, where the numerator and denominator are zero or undefined, and determine the intervals where the expression is non-negative (greater than or equal to zero).

### Step 1: Find the critical points
The critical points occur when either the numerator is zero or the denominator is undefined:
1. \(x + 2 = 0 \) ⇒ \(x = -2\)
2. \(x - 7 = 0 \) ⇒ \(x = 7\)
3. \(x + 3 = 0 \) ⇒ \(x = -3\)

### Step 2: Analyze the sign in different intervals
The critical points divide the number line into the following intervals: \((-∞, -3)\), \((-3, -2)\), \((-2, 7)\), and \((7, ∞)\).

#### Interval \((-∞, -3)\)
- Pick a test point \(x = -4\):
\[
\frac{(x + 2)(x - 7)}{(x + 3)^2} = \frac{(-4 + 2)(-4 - 7)}{(-4 + 3)^2} = \frac{(-2)(-11)}{(-1)^2} = \frac{22}{1} > 0
\]
So, the expression is positive in this interval.

#### Interval \((-3, -2)\)
- Pick a test point \(x = -2.5\):
\[
\frac{(x + 2)(x - 7)}{(x + 3)^2} = \frac{(-2.5 + 2)(-2.5 - 7)}{(-2.5 + 3)^2} = \frac{(-0.5)(-9.5)}{(0.5)^2} = \frac{4.75}{0.25} > 0
\]
So, the expression is positive in this interval.

#### Interval \((-2, 7)\)
- Pick a test point \(x = 0\):
\[
\frac{(x + 2)(x - 7)}{(x + 3)^2} = \frac{(0 + 2)(0 - 7)}{(0 + 3)^2} = \frac{(2)(-7)}{(3)^2} = \frac{-14}{9} < 0
\]
So, the expression is negative in this interval.

#### Interval \((7, ∞)\)
- Pick a test point \(x = 8\):
\[
\frac{(x + 2)(x - 7)}{(x + 3)^2} = \frac{(8 + 2)(8 - 7)}{(8 + 3)^2} = \frac{(10)(1)}{(11)^2} > 0
\]
So, the expression is positive in this interval.

### Step 3: Check the critical points
- At \(x = -2\), the numerator is zero, so the expression equals 0, which satisfies the inequality.
- At \(x = 7\), the numerator is zero, so the expression equals 0, which satisfies the inequality.
- At \(x = -3\), the denominator is zero, so the expression is undefined, meaning \(x = -3\) cannot be included in the solution set.

### Conclusion
The solution set is the union of intervals where the expression is non-negative:

\[
(-\infty, -3) \cup (-2, 7] \cup (7, \infty)
\]

However, checking the options:

- Option (a): \((-∞, -3) \cup (-3, -2) \cup (7, ∞)\)
- Option (b): \((-∞, -3) \cup (7, ∞)\)
- Option (c): \((-3, -2)\)
- Option (d): \((7, ∞)\)
- Option (e): None

Option **(b)** is the correct one, as it matches the intervals where the expression is non-negative.

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Do you have any questions or need further details? Here are 5 related questions to expand your understanding:

1. How do we determine the sign of a rational function in different intervals?
2. What role do critical points play in solving inequalities involving rational expressions?
3. How do we handle the points where the denominator is zero in a rational inequality?
4. Can we solve similar inequalities involving quadratic expressions in the numerator and denominator?
5. How does the concept of sign change apply to polynomial inequalities?

**Tip:** When analyzing rational inequalities, always remember to check the sign of the expression in each interval between critical points, and pay special attention to points where the expression is undefined.

In the equation (x+2)(x-7)/(x+3)^2 >= 0, what will be the value of x?

Learn how to solve the rational inequality (x+2)(x-7)/(x+3)^2 ≥ 0 using critical points and interval testing. This algebra problem is suitable for students in Grades 9-11.

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