1. **State the problem:** Solve the complex equation $$z^3 + 2z^2 + (8 + 6i)z + 16 + 12i = 0$$ given that $$z = -2$$ is a root. 2. **Use the factor theorem:** Since $$-2$$ is a root, $$z + 2$$ is a factor of the polynomial. 3. **Divide the polynomial by $$z + 2$$:** Use polynomial division or synthetic division to find the quadratic factor. 4. **Perform polynomial division:** $$\begin{aligned} &z^3 + 2z^2 + (8 + 6i)z + 16 + 12i \\ = &(z + 2)(Az^2 + Bz + C) \end{aligned}$$ Expand the right side: $$z(Az^2 + Bz + C) + 2(Az^2 + Bz + C) = Az^3 + Bz^2 + Cz + 2Az^2 + 2Bz + 2C$$ Group terms: $$Az^3 + (B + 2A)z^2 + (C + 2B)z + 2C$$ Match coefficients with the original polynomial: $$\begin{cases} A = 1 \\ B + 2A = 2 \\ C + 2B = 8 + 6i \\ 2C = 16 + 12i \end{cases}$$ 5. **Solve the system:** - From $$A = 1$$ - $$B + 2(1) = 2 \Rightarrow B = 0$$ - $$2C = 16 + 12i \Rightarrow C = 8 + 6i$$ - Check $$C + 2B = 8 + 6i + 0 = 8 + 6i$$ correct. 6. **Quadratic factor:** $$z^2 + 0z + (8 + 6i) = 0$$ or $$z^2 + 8 + 6i = 0$$ 7. **Solve quadratic:** $$z^2 = -8 - 6i$$ 8. **Express $$z$$:** $$z = \pm \sqrt{-8 - 6i}$$ 9. **Find square roots of a complex number:** Let $$z = x + yi$$, then $$z^2 = (x + yi)^2 = x^2 - y^2 + 2xyi = -8 - 6i$$ Equate real and imaginary parts: $$\begin{cases} x^2 - y^2 = -8 \\ 2xy = -6 \end{cases}$$ 10. **From $$2xy = -6$$:** $$xy = -3$$ 11. **Express $$y$$ in terms of $$x$$:** $$y = \frac{-3}{x}$$ 12. **Substitute into real part:** $$x^2 - \left(\frac{-3}{x}\right)^2 = -8$$ $$x^2 - \frac{9}{x^2} = -8$$ Multiply both sides by $$x^2$$: $$x^4 - 9 = -8x^2$$ Rewrite: $$x^4 + 8x^2 - 9 = 0$$ 13. **Let $$t = x^2$$:** $$t^2 + 8t - 9 = 0$$ 14. **Solve quadratic in $$t$$:** $$t = \frac{-8 \pm \sqrt{64 + 36}}{2} = \frac{-8 \pm 10}{2}$$ 15. **Calculate roots:** - $$t_1 = \frac{-8 + 10}{2} = 1$$ - $$t_2 = \frac{-8 - 10}{2} = -9$$ (discard since $$x^2$$ cannot be negative for real $$x$$) 16. **So $$x^2 = 1$$, thus $$x = \pm 1$$. 17. **Find corresponding $$y$$:** - For $$x = 1$$, $$y = -3/1 = -3$$ - For $$x = -1$$, $$y = -3/(-1) = 3$$ 18. **Solutions for $$z$$:** $$z = 1 - 3i$$ or $$z = -1 + 3i$$ 19. **Final answer:** The three roots of the polynomial are: $$\boxed{-2, 1 - 3i, -1 + 3i}$$