1. **State the problem:** Find the roots or analyze the cubic polynomial $$x^3 + 2x^2 + 3x + 4$$. 2. **Formula and rules:** For a cubic polynomial $$ax^3 + bx^2 + cx + d = 0$$, roots can be found using factoring, synthetic division, or the cubic formula. Here, $$a=1, b=2, c=3, d=4$$. 3. **Check for rational roots using Rational Root Theorem:** Possible roots are factors of $$d$$ over factors of $$a$$, i.e., $$\pm1, \pm2, \pm4$$. 4. **Test roots:** - For $$x=1$$: $$1 + 2 + 3 + 4 = 10 \neq 0$$ - For $$x=-1$$: $$-1 + 2 - 3 + 4 = 2 \neq 0$$ - For $$x=2$$: $$8 + 8 + 6 + 4 = 26 \neq 0$$ - For $$x=-2$$: $$-8 + 8 - 6 + 4 = -2 \neq 0$$ - For $$x=4$$: $$64 + 32 + 12 + 4 = 112 \neq 0$$ - For $$x=-4$$: $$-64 + 32 - 12 + 4 = -40 \neq 0$$ No rational roots found. 5. **Use depressed cubic substitution or numerical methods:** Since no rational roots, use the cubic formula or approximate roots numerically. 6. **Summary:** The polynomial $$x^3 + 2x^2 + 3x + 4$$ has no rational roots; roots are complex or irrational. Final answer: No rational roots; roots must be found using the cubic formula or numerical approximation.