1. **State the problem:** Solve the equation $$Z^{5} + Z^{3} + Z = 0$$ for $Z$. 2. **Use the factoring technique:** Notice that each term contains a factor of $Z$, so factor it out: $$Z(Z^{4} + Z^{2} + 1) = 0$$ 3. **Apply the zero product property:** For the product to be zero, either $$Z = 0$$ or $$Z^{4} + Z^{2} + 1 = 0$$ 4. **Solve the quartic equation:** Let $x = Z^{2}$, then the equation becomes $$x^{2} + x + 1 = 0$$ 5. **Use the quadratic formula:** $$x = \frac{-1 \pm \sqrt{1^{2} - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{-1 \pm \sqrt{-3}}{2} = \frac{-1 \pm i\sqrt{3}}{2}$$ 6. **Recall $x = Z^{2}$:** So $$Z^{2} = \frac{-1 \pm i\sqrt{3}}{2}$$ 7. **Find $Z$ by taking square roots:** Each complex number has two square roots, so there are four complex roots from this step. 8. **Summary of solutions:** - One real root: $Z = 0$ - Four complex roots from solving $Z^{2} = \frac{-1 \pm i\sqrt{3}}{2}$ Thus, the equation has 5 roots in total, consistent with the degree of the polynomial. **Final answer:** $$Z = 0 \quad \text{or} \quad Z^{2} = \frac{-1 \pm i\sqrt{3}}{2}$$