(x ^ 2 - 1) * (x + 3) ^ 2 >= 0

To solve the inequality \((x^2 - 1)(x + 3)^2 \geq 0\), we will first find the critical points by identifying the values of \(x\) that make the expression equal to zero or undefined. ### Step 1: Find the critical points 1. Set each factor in the expression to zero: - For \(x^2 - 1 = 0\): \[ x^2 = 1 \implies x = 1 \text{ or } x = -1 \] - For \((x + 3)^2 = 0\): \[ x + 3 = 0 \implies x = -3 \] So, the critical points are \(x = -3\), \(x = -1\), and \(x = 1\). ### Step 2: Determine the sign of the expression in the intervals The critical points divide the number line into four intervals: 1. \( (-\infty, -3) \) 2. \( (-3, -1) \) 3. \( (-1, 1) \) 4. \( (1, \infty) \) We will test a point in each interval to determine the sign of the expression. - **Interval \( (-\infty, -3) \)**: Let's pick \(x = -4\) \[ (-4)^2 - 1 = 16 - 1 = 15 > 0 \] \[ (-4 + 3)^2 = (-1)^2 = 1 > 0 \quad \Rightarrow \quad (15)(1) > 0 \] Thus, the expression is **positive** in this interval. - **Interval \( (-3, -1) \)**: Let's pick \(x = -2\) \[ (-2)^2 - 1 = 4 - 1 = 3 > 0 \] \[ (-2 + 3)^2 = (1)^2 = 1 > 0 \quad \Rightarrow \quad (3)(1) > 0 \] Thus, the expression is **positive** in this interval. - **Interval \( (-1, 1) \)**: Let's pick \(x = 0\) \[ (0)^2 - 1 = 0 - 1 = -1 < 0 \] \[ (0 + 3)^2 = (3)^2 = 9 > 0 \quad \Rightarrow \quad (-1)(9) < 0 \] Thus, the expression is **negative** in this interval. - **Interval \( (1, \infty) \)**: Let's pick \(x = 2\) \[ (2)^2 - 1 = 4 - 1 = 3 > 0 \] \[ (2 + 3)^2 = (5)^2 = 25 > 0 \quad \Rightarrow \quad (3)(25) > 0 \] Thus, the expression is **positive** in this interval. ### Step 3: Include the critical points Next, we check the sign at the critical points: - At \(x = -3\): \[ (x^2 - 1)(x + 3)^2 = (9 - 1)(0) = 0 \] - At \(x = -1\): \[ (x^2 - 1)(x + 3)^2 = (1 - 1)(4) = 0 \] - At \(x = 1\): \[ (x^2 - 1)(x + 3)^2 = (1 - 1)(16) = 0 \] ### Step 4: Combine the results The expression \((x^2 - 1)(x + 3)^2 \geq 0\) is satisfied in: - The intervals \( (-\infty, -3) \), \( (-3, -1) \), and \( (1, \infty) \) - At the points \(x = -3\), \(x = -1\), and \(x = 1\) ### Final Solution Combining all the information, the solution to the inequality \((x^2 - 1)(x + 3)^2 \geq 0\) is: \[ (-\infty, -3] \cup [-1, 1] \cup [1, \infty) \]