Solve the following inequality. \[ (x-7)(x-1)^{2}(x+4)<0 \]

To solve the inequality \((x-7)(x-1)^{2}(x+4)<0\), we first find the roots of the corresponding equation \((x-7)(x-1)^{2}(x+4) = 0\). The roots are: - \(x = 7\) - \(x = 1\) (with multiplicity 2) - \(x = -4\) These roots divide the number line into the following intervals: 1. \((- \infty, -4)\) 2. \((-4, 1)\) 3. \((1, 7)\) 4. \((7, +\infty)\) Next, we test each interval to see where the product \((x-7)(x-1)^{2}(x+4)\) is negative. 1. **Interval \((- \infty, -4)\)**: Choose \(x = -5\): \[ (-5-7)(-5-1)^{2}(-5+4) = (-12)(36)(-1) > 0 \quad (\text{positive}) \] 2. **Interval \((-4, 1)\)**: Choose \(x = 0\): \[ (0-7)(0-1)^{2}(0+4) = (-7)(1)(4) < 0 \quad (\text{negative}) \] 3. **Interval \((1, 7)\)**: Choose \(x = 2\): \[ (2-7)(2-1)^{2}(2+4) = (-5)(1)(6) < 0 \quad (\text{negative}) \] 4. **Interval \((7, +\infty)\)**: Choose \(x = 8\): \[ (8-7)(8-1)^{2}(8+4) = (1)(49)(12) > 0 \quad (\text{positive}) \] Now, we summarize what we found: - \((- \infty, -4)\): positive - \((-4, 1)\): negative - \((1, 7)\): negative - \((7, +\infty)\): positive Next, we check the critical points: - At \(x = -4\): \(((-4-7)(-4-1)^{2}(-4+4) = 0)\) - At \(x = 1\): \((1-7)(1-1)^{2}(1+4) = 0)\) - At \(x = 7\): \((7-7)(7-1)^{2}(7+4) = 0)\) Since we are looking for where \((x-7)(x-1)^{2}(x+4)<0\), we can conclude that the solution to the inequality is: \[ \boxed{(-4, 1) \cup (1, 7)} \]