One state space representation of $G(z)$ is,
$$
\begin{array}{l c r}
A = \begin{bmatrix}
0 & 0 & -1 \\
1 & 0 & 5 \\
0 & 1 & 1
\end{bmatrix} &
B = \begin{bmatrix}
1 \\ 2 \\ 1
\end{bmatrix} &
C = \begin{bmatrix}
0 & 0 & 1
\end{bmatrix}
\end{array}
$$
with dynamics,
$$
x(t+1) = A\,x(t) + B\,u(t)
$$
$$
y(t) = C\,x(t)
$$
You want to find a $L$ and $K$, such that both $A-L\,C$ and $A-B\,K$ have of its eigenvalues at zero. Since I already put the system in the observable canonical form, finding $L$ will be easy, namely the last column of $A$. In order to find $K$ it might be easier to transform the system to the controllable canonical form, find $K$ there and transform it back.
You can now estimate the internal states of the system $x(t)$ with $\hat{x}(t)$, using information from $y(t)$, $u(t)$ and the dynamics of the system,
$$
\hat{x}(t+1) = L\,y(t) + \left(A - L\,C\right) \hat{x}(t) + B\,u(t)
$$
If you choose the input $u(t)$ as follows,
$$
u(t) = -K\,\hat{x}(t)
$$
and define $e(t)$ as the error between the actual state $x(t)$ and its prediction $\hat{x}(t)$,
$$
e(t) = x(t) - \hat{x}(t)
$$
then the total dynamics of the system can be written as,
$$
\begin{bmatrix}
x(t+1) \\ e(t+1)
\end{bmatrix} =
\begin{bmatrix}
A - B\,K & B\,K \\
0 & A - L\,C
\end{bmatrix}
\begin{bmatrix}
x(t) \\ e(t)
\end{bmatrix}
$$
However you have chosen the eigenvalues of each matrix on the diagonal to have all zero eigenvalues, so after six step the system will always go to zero (independent of how good your initial guess is for $\hat{x}(t)$).
