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:
$$
\begin{pmatrix}
1 & 0 & 0 & 0\\
1 & 1 & 0 & 0\\
0 & 1 & 1 & 0\\
0 & 0 & 1 & -1\\
\end{pmatrix}
$$

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
$$
\begin{pmatrix}
1  &  0  &  0  \\
1  &  1  &  0  \\
0  &  1  &  1
\end{pmatrix}
$$
