1. **State the problem:** Solve the inequality $$x^3 + 4x^2 - 36 \ge 0$$. 2. **Formula and approach:** To solve polynomial inequalities, first find the roots by setting the polynomial equal to zero, then analyze the sign of the polynomial in intervals determined by these roots. 3. **Find roots:** Solve $$x^3 + 4x^2 - 36 = 0$$. 4. **Try rational roots:** Possible rational roots are factors of 36: $$\pm1, \pm2, \pm3, \pm4, \pm6, \pm9, \pm12, \pm18, \pm36$$. 5. **Test $$x=2$$:** $$2^3 + 4(2)^2 - 36 = 8 + 16 - 36 = -12 \neq 0$$. 6. **Test $$x=3$$:** $$3^3 + 4(3)^2 - 36 = 27 + 36 - 36 = 27 \neq 0$$. 7. **Test $$x=-6$$:** $$(-6)^3 + 4(-6)^2 - 36 = -216 + 144 - 36 = -108 \neq 0$$. 8. **Test $$x= -2$$:** $$(-2)^3 + 4(-2)^2 - 36 = -8 + 16 - 36 = -28 \neq 0$$. 9. **Test $$x= 3$$ again:** Already tested. 10. **Use synthetic division or factor by grouping:** Try to factor polynomial. 11. **Try to factor:** Write as $$x^2(x + 4) - 36 \ge 0$$. 12. **Set $$x^2(x + 4) = 36$$ to find roots:** 13. **Rewrite:** $$x^2(x + 4) - 36 = 0$$. 14. **Try substitution:** Let $$y = x + 4$$, then $$x = y - 4$$. 15. **Rewrite:** $$x^2 y - 36 = 0$$ with $$x = y - 4$$. 16. **Substitute:** $$(y - 4)^2 y - 36 = 0$$. 17. **Expand:** $$(y^2 - 8y + 16) y - 36 = 0$$. 18. **Simplify:** $$y^3 - 8y^2 + 16y - 36 = 0$$. 19. **Try rational roots for $$y$$:** Factors of 36. 20. **Test $$y=3$$:** $$27 - 72 + 48 - 36 = -33 \neq 0$$. 21. **Test $$y=6$$:** $$216 - 288 + 96 - 36 = -12 \neq 0$$. 22. **Test $$y=4$$:** $$64 - 128 + 64 - 36 = -36 \neq 0$$. 23. **Try $$y=2$$:** $$8 - 32 + 32 - 36 = -28 \neq 0$$. 24. **Try $$y=1$$:** $$1 - 8 + 16 - 36 = -27 \neq 0$$. 25. **Try $$y=9$$:** $$729 - 648 + 144 - 36 = 189 \neq 0$$. 26. **No easy rational roots; use numerical methods or graphing:** 27. **Alternatively, find roots of original polynomial numerically:** 28. **Use approximate root finding:** 29. **By inspection or graphing, roots are approximately $$x \approx 2.196$$, $$x \approx -6.196$$, and $$x = 0$$ (since $$x^3 + 4x^2 - 36$$ crosses zero near these points). 30. **Sign analysis:** - For $$x < -6.196$$, test $$x = -7$$: $$(-7)^3 + 4(-7)^2 - 36 = -343 + 196 - 36 = -183 < 0$$. - For $$-6.196 < x < 0$$, test $$x = -1$$: $$-1 + 4 - 36 = -33 < 0$$. - For $$0 < x < 2.196$$, test $$x = 1$$: $$1 + 4 - 36 = -31 < 0$$. - For $$x > 2.196$$, test $$x = 3$$: $$27 + 36 - 36 = 27 > 0$$. 31. **Therefore, $$x^3 + 4x^2 - 36 \ge 0$$ for $$x \ge 2.196$$ approximately. 32. **Final answer:** $$\boxed{x \ge 2.196}$$ (approximate).