I have the following information:
$p_A(\lambda) = (\lambda - 1)^6(\lambda + 2)^4$ and $m_A(\lambda) = (\lambda - 1)^3(\lambda + 2)^2$
and also $\dim(\operatorname{Ker}(A-I)^2) = 5$ and $\dim(\operatorname{Ker}(A + 2I)) = 2$
I'm trying to find all Jordan forms.
I know form the characteristic polynomial that the matrix will be $10 \times 10$ and that the eigenvalues are $1$ and $-2$.
From $m_A(\lambda) = (\lambda - 1)^3(\lambda + 2)^2$ and $\dim(\operatorname{Ker}(A + 2I)) = 2$, we know that there will be two $2 \times 2$ Jordan blocks for eigenvalue $-2$.
The largest Jordan block for $1$ is $ 3 \times 3$. I'm unsure how to proceed here. I don't understand what information $\dim(\operatorname{Ker}(A-I)^2) = 5$ gives me, or in general what effect $\dim(\operatorname{Ker}(A- \lambda I)^n) = m$ has on the Jordan form.
Your statement "From $m_A(\lambda) = (\lambda - 1)^3(\lambda + 2)^2$ and $\dim(\operatorname{Ker}(A + 2I)) = 2$, we know that there will be two $2 \times 2$ Jordan blocks for eigenvalue $-2$ is not correct. For example the matrix
$$B=\begin{pmatrix} -2 & 0 & 0\\ 0 & -2 & 1\\ 0 & 0 & -2 \end{pmatrix}$$ is such that $\dim \ker(B+2I) = 2$ and $(B+2I)^2=0$. However, the equality $\dim \ker (A+2I) = 2$, the fact that $m_A(x) = (x-1)^3(x+2)^2$ and the characteristic polynomial allow us to conclude that either $A$ has two Jordan blocks of size 2 or one Jordan block of size two and two of size one for the eigenvalue $-2$.
Regarding the eigenvalue $1$, denote by $n_1,n_2,n_3$ the Jordan blocks of size $1,2,3$ respectively. From the minimal polynomial, we know that $n_3 \ge 1$. From the characteristic polynomial, we know that $n_1+2n_2+3n_3 = 6$. We also have $n_1+2n_2+2n_3= \dim \ker (A-I)^2=5$. Therefore $n_3=1$ and $n_1+2n_2=3$. Which implies finally
$$(n_1,n_2,n_3) \in \{(3,0,1), (1,1,1)\}$$
And we're done.