1. **State the problem:** We need to determine the stability of the discrete-time system with characteristic polynomial $$p(z) = z^6 + 2z^5 + 3z^4 + 5z^3 + 18z^3 + 4z + 16$$ using the Jury stability test. 2. **Simplify the polynomial:** Notice the polynomial has two $z^3$ terms. Combine them: $$p(z) = z^6 + 2z^5 + 3z^4 + (5 + 18)z^3 + 4z + 16 = z^6 + 2z^5 + 3z^4 + 23z^3 + 4z + 16$$ 3. **Write the polynomial in standard form:** $$p(z) = a_0 z^6 + a_1 z^5 + a_2 z^4 + a_3 z^3 + a_4 z^2 + a_5 z + a_6$$ From the polynomial, coefficients are: $$a_0 = 1, a_1 = 2, a_2 = 3, a_3 = 23, a_4 = 0, a_5 = 4, a_6 = 16$$ 4. **Jury test conditions:** For a polynomial of degree $n$, the system is stable if all roots lie inside the unit circle. The Jury test involves checking the following conditions: - $|a_6| < a_0$ (leading and constant term condition) - $|a_0| > 0$ - Construct the Jury table and check positivity of certain determinants. 5. **Check first condition:** $$|a_6| = |16| = 16$$ $$a_0 = 1$$ Since $16 \not< 1$, the first condition fails. 6. **Conclusion:** Since the first Jury condition $|a_6| < a_0$ is not satisfied, the system is **not stable**. **Final answer:** The system with polynomial $$p(z) = z^6 + 2z^5 + 3z^4 + 23z^3 + 4z + 16$$ is **not stable** according to the Jury stability test because $|a_6|$ is not less than $a_0$.