We are tasked with solving the inequality:

$(x + 9)^2(x + 3)(x - 1) > 0$

Step 1: Identify critical points

The critical points are the values of $x$ where the expression equals zero. To find these points, set each factor in the expression to zero:

1. $(x + 9)^2 = 0 \Rightarrow x = -9$
2. $(x + 3) = 0 \Rightarrow x = -3$
3. $(x - 1) = 0 \Rightarrow x = 1$

Thus, the critical points are $x = -9$, $x = -3$, and $x = 1$.

Step 2: Analyze the intervals

The critical points divide the real number line into four intervals:

1. $(-\infty, -9)$
2. $(-9, -3)$
3. $(-3, 1)$
4. $(1, \infty)$

We now analyze the sign of the expression $(x + 9)^2(x + 3)(x - 1)$ on each interval.

Interval 1: $(-\infty, -9)$

• Choose $x = -10$:

  • $(x + 9)^2 = (-10 + 9)^2 = 1$ (positive)
  • $(x + 3) = (-10 + 3) = -7$ (negative)
  • $(x - 1) = (-10 - 1) = -11$ (negative)

  The product is positive: $(+)(-)(-) = +$.

Interval 2: $(-9, -3)$

• Choose $x = -5$:

  • $(x + 9)^2 = (-5 + 9)^2 = 16$ (positive)
  • $(x + 3) = (-5 + 3) = -2$ (negative)
  • $(x - 1) = (-5 - 1) = -6$ (negative)

  The product is positive: $(+)(-)(-) = +$.

Interval 3: $(-3, 1)$

• Choose $x = 0$:

  • $(x + 9)^2 = (0 + 9)^2 = 81$ (positive)
  • $(x + 3) = (0 + 3) = 3$ (positive)
  • $(x - 1) = (0 - 1) = -1$ (negative)

  The product is negative: $(+)(+)(-) = -$.

Interval 4: $(1, \infty)$

• Choose $x = 2$:

  • $(x + 9)^2 = (2 + 9)^2 = 121$ (positive)
  • $(x + 3) = (2 + 3) = 5$ (positive)
  • $(x - 1) = (2 - 1) = 1$ (positive)

  The product is positive: $(+)(+)(+) = +$.

Step 3: Consider the critical points

• At $x = -9$, $(x + 9)^2 = 0$, so the expression is zero.
• At $x = -3$ and $x = 1$, the expression is zero because of the $(x + 3)$ and $(x - 1)$ factors, respectively.

Step 4: Solution in interval notation

The expression is positive on the intervals where the product is positive. These intervals are:

• $(-\infty, -9)$
• $(-9, -3)$
• $(1, \infty)$

Since we want $(x + 9)^2(x + 3)(x - 1) > 0$, we exclude the points where the expression equals zero.

Thus, the solution is:

$(-\infty, -9) \cup (-9, -3) \cup (1, \infty)$

Would you like any more details or clarifications?

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Here are 5 related questions:

1. How would the solution change if the inequality were $\geq 0$?
2. What happens if you replace one of the linear factors with a higher-degree term?
3. How do we handle inequalities with rational expressions?
4. What is the role of multiplicity in determining the sign of intervals?
5. How does squaring a factor like $(x + 9)$ affect the sign analysis?

Tip: When solving inequalities, always remember to test values from each interval to check the sign of the expression.