Find the Jordan Canonical Form from c(x) and m(x) Given the matrix B:
$$
\begin{pmatrix}
2 & 1 & -2 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
1 & 0 & 0 & 0 \\
\end{pmatrix}
$$
I have found that the characteristic polynomial is:
$
c(x)=(x-1)^{3}(x+1)
$
and then found that the minimal polynomial is:
$
m(x)=c(x)
$
so then can I conclude that  is similar to the Jordan Canonical matrix:
$$
\begin{pmatrix}
1 & 0 & 0 & 0\\
1 & 1 & 0 & 0\\
0 & 1 & 1 & 0\\
0 & 0 & 1 & -1\\
\end{pmatrix}
$$
 A: From the comment of N.S., and CW. 
$$
\left(\begin{matrix}
1 & 0 & 0 & 0\\
1 & 1 & 0 & 0\\
0 & 1 & 1 & 0\\
0 & 0 & 0 & -1\\
\end{matrix}\right)
$$ 
Note that, despite the multiple eigenvalue, any matrix that commutes with $B$ can be written as a polynomial $p(B).$ 
A: I'm not sure how you computed the minimal polynomial, but it's not a necessary step -nor its knowledge will give you, in general, the Jordan canonical form.
Instead, once I had the characteristic polynomial, I would procede as follows. Assume it looks like
$$
Q(t) = (t-\lambda)^m \dots
$$
That is: the eigenvalue $\lambda$ has algebraic multiplicity equal to $m$. That is, the Jordan block of $\lambda$ has size $m\times m$.
Then, I would compute the following ranks:
$$
\mathrm{rank} (A - \lambda I) > \mathrm{rank} (A - \lambda I)^2 > \mathrm{rank} (A - \lambda I)^3 > \dots
$$
Or, what is the same, the dimensions of the following subspaces:
$$
\mathrm{Ker} (A - \lambda I) \subset \mathrm{Ker} (A - \lambda I)^2 \subset \mathrm{Ker} (A - \lambda I)^3 \subset \dots 
$$
When I got
$$
d_\alpha = \mathrm{dim}\ \mathrm{Ker} (A - \lambda )^\alpha = m
$$
I would stop. All the information you need about the Jordan canonical form is contained in the sequence of numbers
$$
d_i = \mathrm{dim}\ \mathrm{Ker} (A - \lambda I )^i \ , \qquad i = 1, \dots, \alpha \ .
$$
For instance,
$$
d_1 = \mathrm{dim}\ \mathrm{Ker} (A - \lambda I ) = n - \mathrm{rank} (A - \lambda I)
$$
is the number of "boxes" inside the Jordan block of the eigenvalue $\lambda$. (Here, $n$ is the size of your matrix.)
In your example, 
$$
d_1 = \mathrm{dim}\ \mathrm{Ker} (A - I ) = 4 - \mathrm{rank} (A - I) = 4- 3 = 1 
$$
means there is just one box in the Jordan block of $1$. So, it must be $3 \times 3$. That is, it must be
$$
\begin{pmatrix}
1  &  0  &  0  \\
1  &  1  &  0  \\
0  &  1  &  1
\end{pmatrix}
$$
A: I am not sure what you really are asking for, but here is a result you might find useful. As Arturo Magidin proved here:

The minimal polynomial is equal to the characteristic polynomial if and only if the dimension of each eigenspace is 1. 

What can you do with this? Suppose the (distinct) eigenvalues of a matrix $A$ are $\lambda_1,\ldots,\lambda_r$, and suppose that each eigenspace is equal to 1. That is, the geometric multiplicity of each eigenvalue is 1. This means that there are exactly $r$ Jordan blocks in the Jordan canonical form, one for each eigenvalue. Therefore, the Jordan canonical form is
$$\left(\begin{matrix}J_1& &\\ &\ddots& \\ & &J_r\end{matrix}\right),$$
where $J_i$ is a Jordan block of size $m_i\times m_i$ ($m_i$ is the algebraic multiplicity of $\lambda_i$) for the eigenvalue $\lambda_i$. In particular, the Jordan canonical form for your matrix $B$ is the one in Will Jagy's answer.
