1. **State the problem:** Solve the inequality $$\frac{x^2 - 2x - 8}{x^3 + x} \geq 0$$. 2. **Factor numerator and denominator:** - Numerator: $$x^2 - 2x - 8 = (x - 4)(x + 2)$$ - Denominator: $$x^3 + x = x(x^2 + 1)$$ 3. **Analyze denominator:** - Since $$x^2 + 1 > 0$$ for all real $$x$$, the sign of the denominator depends only on $$x$$. 4. **Rewrite inequality:** $$\frac{(x - 4)(x + 2)}{x(x^2 + 1)} \geq 0 \implies \frac{(x - 4)(x + 2)}{x} \geq 0$$ 5. **Find critical points:** - Numerator zeros: $$x = 4, -2$$ - Denominator zero: $$x = 0$$ (excluded from domain) 6. **Determine sign intervals:** - Intervals: $$(-\infty, -2), (-2, 0), (0, 4), (4, \infty)$$ 7. **Test each interval:** - For $$x < -2$$, choose $$x = -3$$: $$\frac{(-3 - 4)(-3 + 2)}{-3} = \frac{(-7)(-1)}{-3} = \frac{7}{-3} < 0$$ - For $$-2 < x < 0$$, choose $$x = -1$$: $$\frac{(-1 - 4)(-1 + 2)}{-1} = \frac{(-5)(1)}{-1} = \frac{-5}{-1} = 5 > 0$$ - For $$0 < x < 4$$, choose $$x = 1$$: $$\frac{(1 - 4)(1 + 2)}{1} = (-3)(3) = -9 < 0$$ - For $$x > 4$$, choose $$x = 5$$: $$\frac{(5 - 4)(5 + 2)}{5} = (1)(7)/5 = 7/5 > 0$$ 8. **Include zeros where numerator is zero:** - At $$x = -2$$ and $$x = 4$$, numerator is zero, so expression equals zero. - At $$x = 0$$, denominator zero, undefined. 9. **Solution:** $$[-2, 0) \cup [4, \infty)$$ **Final answer:** $$\boxed{[-2, 0) \cup [4, \infty)}$$