Inequality Solution 9Ce887

1. **State the problem:** Solve the inequality $$(x + 2)(x - 3)(x - 5) \leq 0$$ and express the solution using inequality notation. 2. **Identify critical points:** The expression equals zero when any factor is zero: $$x + 2 = 0 \Rightarrow x = -2$$ $$x - 3 = 0 \Rightarrow x = 3$$ $$x - 5 = 0 \Rightarrow x = 5$$ These points divide the number line into intervals: $$(-\infty, -2), (-2, 3), (3, 5), (5, \infty)$$ 3. **Test each interval:** Choose a test point in each interval to determine the sign of the product. - For $$x = -3$$ in $$(-\infty, -2)$$: $$(x+2)(x-3)(x-5) = (-3+2)(-3-3)(-3-5) = (-1)(-6)(-8) = -48 < 0$$ - For $$x = 0$$ in $$(-2, 3)$$: $$(0+2)(0-3)(0-5) = (2)(-3)(-5) = 30 > 0$$ - For $$x = 4$$ in $$(3, 5)$$: $$(4+2)(4-3)(4-5) = (6)(1)(-1) = -6 < 0$$ - For $$x = 6$$ in $$(5, \infty)$$: $$(6+2)(6-3)(6-5) = (8)(3)(1) = 24 > 0$$ 4. **Determine solution intervals:** The inequality requires the product to be less than or equal to zero, so include intervals where the product is negative or zero. From tests: - Negative on $$(-\infty, -2)$$ and $$(3, 5)$$ - Zero at $$x = -2, 3, 5$$ 5. **Write the solution:** Include points where product equals zero and intervals where product is negative: $$(-\infty, -2] \cup [3, 5]$$ 6. **Final answer:** $$\boxed{(-\infty, -2] \cup [3, 5]}$$ This is the solution set for the inequality $$(x + 2)(x - 3)(x - 5) \leq 0$$.