1. **State the problem:** Solve the cubic equation $$x^3 + 3x^2 - x + 12 = 0$$ and find the real root first, then express the complex roots in the form $$a \pm \sqrt{b}i$$. 2. **Use the Rational Root Theorem** to test possible rational roots among factors of 12: $$\pm1, \pm2, \pm3, \pm4, \pm6, \pm12$$. 3. **Test $x = -3$:** $$(-3)^3 + 3(-3)^2 - (-3) + 12 = -27 + 27 + 3 + 12 = 15 \neq 0$$ 4. **Test $x = -2$:** $$(-2)^3 + 3(-2)^2 - (-2) + 12 = -8 + 12 + 2 + 12 = 18 \neq 0$$ 5. **Test $x = -1$:** $$(-1)^3 + 3(-1)^2 - (-1) + 12 = -1 + 3 + 1 + 12 = 15 \neq 0$$ 6. **Test $x = 1$:** $$1 + 3 - 1 + 12 = 15 \neq 0$$ 7. **Test $x = 2$:** $$8 + 12 - 2 + 12 = 30 \neq 0$$ 8. **Test $x = -4$:** $$(-4)^3 + 3(-4)^2 - (-4) + 12 = -64 + 48 + 4 + 12 = 0$$ So, $x = -4$ is a root. 9. **Divide the cubic by $(x + 4)$** using polynomial division or synthetic division: $$\frac{x^3 + 3x^2 - x + 12}{x + 4} = x^2 - x + 3$$ 10. **Solve the quadratic $x^2 - x + 3 = 0$** using the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{1 \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot 3}}{2} = \frac{1 \pm \sqrt{1 - 12}}{2} = \frac{1 \pm \sqrt{-11}}{2}$$ 11. **Express the complex roots:** $$x = \frac{1}{2} \pm \frac{\sqrt{11}}{2}i$$ **Final answers:** - Real root: $$x = -4$$ - Complex roots: $$x = \frac{1}{2} \pm \frac{\sqrt{11}}{2}i$$