1. Problem statement: Determine stability using the Jury test for the polynomial $p(z)=z^6+2z^5+3z^4+23z^3+0z^2+4z+16$. 2. Recall the Jury test rules: For a polynomial of degree $n$ to have all roots inside the unit circle a necessary condition is $p(1)>0$ and $(-1)^n p(-1)>0$; if these hold we proceed to build the full Jury table to check all conditions. 3. Compute $p(1)$: $p(1)=1+2+3+23+0+4+16=49$, hence $p(1)=49>0$. 4. Compute $p(-1)$: $p(-1)=(-1)^6+2(-1)^5+3(-1)^4+23(-1)^3+0(-1)^2+4(-1)+16$. 5. Simplify the powers and sum: $p(-1)=1-2+3-23+0-4+16=-9$. 6. Since $n=6$ is even we require $p(-1)>0$; here $p(-1)=-9<0$, so this necessary Jury condition fails and the polynomial cannot be stable (it has at least one root outside the unit circle). 7. (Illustration) When starting the Jury table the first elimination ratio would be $\frac{a_0}{a_n}=\frac{16}{16}=1$; show cancellation explicitly: $$\frac{\cancel{16}}{\cancel{16}}=1$$. 8. Final answer: The system is unstable; $p(-1)=-9<0$, so the polynomial does not satisfy the Jury necessary condition for all roots inside the unit circle.