How to turn this matrix to Jordan normal form? Matrix $A$ is $
\left( \begin{array}{ccc}
3 & 0 & 8 \\
3 & -1 & 6 \\
-2 & 0 & -5 \end{array} \right)$ and I need to find a matrix P such that $P^{-1} A P = J$ where $J$ is a Jordan matrix.
My trial is:
Calculate the $det(λI - A) = (λ+1)^{3}$ and its elementary divisor is $(λ+1)^{3}$ as well; 
So get $J = \left( \begin{array}{ccc}
-1 & 1 & 0 \\
0 & -1 & 1 \\
0 & 0 & -1 \end{array} \right)$ being a Jordan matrix of A; 
Then suppose matrix $P$ = $[\vec{x_1} \ \vec{x_2} \ \vec{x_3}]$ and turn $P^{-1} A P = J$ to $AP = A[\vec{x_1} \ \vec{x_2} \ \vec{x_3}] = [A\vec{x_1} \ A\vec{x_2} \ A\vec{x_3}] = PJ = [\vec{x_1} \ \vec{x_2} \ \vec{x_3}] \left( \begin{array}{ccc}
-1 & 1 & 0 \\
0 & -1 & 1 \\
0 & 0 & -1 \end{array} \right)$
And then I got stuck here: I try to solve this equation but no solution for $\vec{x_2}$ and $\vec{x_3}$ at all. Where am I wrong? Please help me find it.
Thanks in advance.
Update:
I'm not sure of my inspection of the single elementary divisor there and so I upload my trail:
$det(λI - A) = \begin{vmatrix}
λ-3 & 0 & -8 \\
-3 & λ+1 & -6 \\
2 & 0 & λ+5 \end{vmatrix}$;
Greatest common divisor of 1st subdeterminants is $1$ obviously;
I'm not sure about greatest common divisor of 2nd subdeterminants and I think it should be $1$ coz $ \begin{vmatrix}
λ-3 & 0 \\
-3 & λ+1 \end{vmatrix}$ and $ \begin{vmatrix}
0 & -8 \\
0 & λ+5 \end{vmatrix}$ are relatively prime. Is there something wrong here? 
Update2:
Brian Fitzpatrick helped me find out the error a lot that my problem is on 2nd subdeterminant $\begin{vmatrix} 0 & -8 \\ 0 & λ+5 \end{vmatrix}$ which should not be picked up and used as a comparison for λ-prime(order of λ should be at least $1$) and then the jordan normal form will be corrected as $ \left(\begin{array}{ccc} -1 & 0 & 0 \\ 0 & -1 & 1 \\ 0 & 0 & -1 \end{array} \right)$.
Besides, the step above where I got stucked does being no solution at all as $ \vec {x_2} = [-2 \ 0 \ 1]$(one eigenvector of A and the other is $[0 \ 1 \ 0]$) and that does not mean you can't find a generalized eigenvector. Here a trick introduced that can help you: just set $ \vec {x_2} = k_1 * [0 \ 1 \ 0] + k_2 * [-2 \ 0 \ 1]$ and then put it in $(A+I) \vec{x_3} = \vec{x_2}$ solving relationship between $k_1$ and $k_2$. And then $P = \left(\begin{array}{ccc} 0 & -4 & 1 \\ 1 & -3 & 0 \\ 0 & 2 & -1 \end{array} \right)$. 
 A: Let's follow the algorithm described here.
The characteristic polynomial of $A$ is
$$
\DeclareMathOperator{char}{char}\char_A(t)=-(t+1)^3
$$
so the only eigenvalue of $A$ is $\lambda=-1$ with algebraic multiplicity $m=3$. Note that
\begin{align*}
A+I&=\begin{bmatrix}4&0&8\\ 3&0&6\\-2&0&-4\end{bmatrix} &
(A+I)^2 &= \begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}
\end{align*}
so $\DeclareMathOperator{rank}{rank}\DeclareMathOperator{null}{null}$
\begin{align*}
\dim\null(A+I) &= 2 & \dim\null(A+I)^2&=3
\end{align*}
We then compute the numbers
\begin{align*}
d_1 &= \dim\null(A+I) & d_2 &= \dim\null(A+I)^2-\dim\null(A+I) \\
&= 2 & &= 3-2 \\
&&&= 1
\end{align*}
so we must fill the boxes
$$
\begin{matrix}
\Box & \Box \\
\Box
\end{matrix}
$$
with vectors. Note that $u=\begin{bmatrix}1&0&0\end{bmatrix}^\top$ satisfies $u\in\null(A+I)^2$ but $u\notin\null(A+I)$. Put $v=(A+I)u=\begin{bmatrix}4&3&-2\end{bmatrix}^\top$ so the diagram takes the form
$$
\begin{matrix}
v & \Box \\
u &
\end{matrix}
$$
Since $w=\begin{bmatrix} 0&1&0\end{bmatrix}^\top$ is linearly independent from $v$ we complete the Jordan basis $\{u,v,w\}$.
Finally, put
$$
P=
\begin{bmatrix}
4&1&0\\ 3&0&1\\ -2&0&0
\end{bmatrix}
$$
and note that the Jordan form is 
$$
J=P^{-1}AP=\begin{bmatrix}-1&1&0\\0&-1&0\\0&0&-1\end{bmatrix}
$$
