1. **State the problem:** Solve the inequality $$(x + 2)(x - 3)(x + 3)(x + \geq 0$$ (assuming the last factor is $(x + 1)$ due to typo). 2. **Rewrite the inequality:** $$ (x + 2)(x - 3)(x + 3)(x + 1) \geq 0 $$ 3. **Identify critical points:** Set each factor equal to zero: - $x + 2 = 0 \Rightarrow x = -2$ - $x - 3 = 0 \Rightarrow x = 3$ - $x + 3 = 0 \Rightarrow x = -3$ - $x + 1 = 0 \Rightarrow x = -1$ 4. **Plot critical points on a number line:** Points: $-3, -2, -1, 3$ 5. **Determine sign of each factor in intervals:** Intervals: $(-\infty, -3), (-3, -2), (-2, -1), (-1, 3), (3, \infty)$ 6. **Test each interval:** - For $x < -3$, pick $x = -4$: - $(x+2) = -2$ (negative) - $(x-3) = -7$ (negative) - $(x+3) = -1$ (negative) - $(x+1) = -3$ (negative) Product of four negatives: $(-)(-)(-)(-) = +$ (positive) - For $-3 < x < -2$, pick $x = -2.5$: - $(x+2) = -0.5$ (negative) - $(x-3) = -5.5$ (negative) - $(x+3) = 0.5$ (positive) - $(x+1) = -1.5$ (negative) Product: $(-)(-)(+)(-) = -$ (negative) - For $-2 < x < -1$, pick $x = -1.5$: - $(x+2) = 0.5$ (positive) - $(x-3) = -4.5$ (negative) - $(x+3) = 1.5$ (positive) - $(x+1) = -0.5$ (negative) Product: $(+)(-)(+)(-) = +$ (positive) - For $-1 < x < 3$, pick $x = 0$: - $(x+2) = 2$ (positive) - $(x-3) = -3$ (negative) - $(x+3) = 3$ (positive) - $(x+1) = 1$ (positive) Product: $(+)(-)(+)(+) = -$ (negative) - For $x > 3$, pick $x = 4$: - $(x+2) = 6$ (positive) - $(x-3) = 1$ (positive) - $(x+3) = 7$ (positive) - $(x+1) = 5$ (positive) Product: $(+)(+)(+)(+) = +$ (positive) 7. **Include points where product is zero:** At $x = -3, -2, -1, 3$, product is zero. 8. **Write solution set:** Intervals where product $\geq 0$ are: $$ (-\infty, -3] \cup [-2, -1] \cup [3, \infty) $$ **Final answer:** $$ \boxed{\{x \mid x \leq -3 \text{ or } -2 \leq x \leq -1 \text{ or } x \geq 3\}} $$