Solve the following inequality. \[ (x+2)^{2}(x-8)>0 \]

To solve the inequality \( (x+2)^{2}(x-8) > 0 \), we first identify the critical points where the expression is equal to zero. These occur at \( x + 2 = 0 \) and \( x - 8 = 0 \), leading to critical points at: 1. \( x = -2 \) 2. \( x = 8 \) Next, we analyze the sign of the expression in the intervals determined by these critical points: 1. \( (-\infty, -2) \) 2. \( (-2, 8) \) 3. \( (8, \infty) \) Choose test points from each interval: - **Interval \( (-\infty, -2) \)**: Let \( x = -3 \) \[ (-3 + 2)^{2}(-3 - 8) = (-1)^{2}(-11) = 1 \cdot (-11) = -11 < 0 \] - **Interval \( (-2, 8) \)**: Let \( x = 0 \) \[ (0 + 2)^{2}(0 - 8) = (2)^{2}(-8) = 4 \cdot (-8) = -32 < 0 \] - **Interval \( (8, \infty) \)**: Let \( x = 9 \) \[ (9 + 2)^{2}(9 - 8) = (11)^{2}(1) = 121 \cdot 1 = 121 > 0 \] Now, we summarize the results: - In \( (-\infty, -2) \), the expression is negative. - In \( (-2, 8) \), the expression is negative. - In \( (8, \infty) \), the expression is positive. Additionally, check the behavior at the critical points: - At \( x = -2 \), \( (x + 2)^{2} = 0 \), so the whole expression equals \( 0 \). - At \( x = 8 \), \( (x - 8) = 0 \), so again the entire product equals \( 0 \). Since we want the inequality \( (x+2)^2(x-8) > 0\), it is satisfied in the interval \( (8, \infty) \) only. Thus, the solution to the inequality is: \[ \boxed{(8, \infty)} \]