Algebraic multiplicity Give the matrix A:
$$
    \begin{matrix}
    2 & 1 & 0 \\
    0 & 0 & 2 \\
    0 & 3 & -1 \\
    \end{matrix}
$$
How can I find the algebraic multiplicity of this polynomial?
$-λ^3 + λ^2 + 8λ - 12$
Knowing that the eigenvalues of the matrix are $2, 2, -3$?
I know I have to substitute the eigenvalues in the polynomial but I can't understand what is the algebraic multiplicity.
How do I find the base for each eigenspace?
 A: I don't think you're looking for the algebraic multiplicity of the polynomial, but of its roots; no?

How can I find the algebraic multiplicity of this polynomial?
$-λ^3 + λ^2 + 8λ - 12$
Knowing that the eigenvalues of the matrix are $2, 2, -3$?

And if this is given, then you already know the multiplicities... If not, it is precisely because:
$$-λ^3 + λ^2 + 8λ - 12 = -\left( \lambda-\color{green}{2} \right)^\color{blue}{2} \left(\lambda-(\color{red}{-3})\right)^\color{purple}{1}$$
that we say that the root $\lambda = \color{green}{2}$ has algebraic multiplicity $\color{blue}{2}$ and the root $\lambda = \color{red}{-3}$ has algebraic multiplicity $\color{purple}{1}$.
In general, we say that $x=\color{green}{a}$ is a root with algebraic multiplicity $\color{blue}{k}$ of a polynomial $p(x)$, if there exists a polynomial $q(x)$ such that $p(x) = (x-\color{green}{a})^\color{blue}{k}q(x)$ with $q(\color{green}{a}) \ne 0$.

Added after comment.

I don't know how to transform $-λ^3 + λ^2 + 8λ - 12$ into a polynomial with brackets

That's a different question; you should look up how to factor a polynomial. It's possible you can avoid having to do this by simplifying the determinant $\det\left(A-\lambda I_n\right)$ in a different way. If you do end up with this cubic polynomial; you can use methods like Horner's rule.
Now that you've given $A$, it is indeed easy to avoid having to factor a cubic polynomial to find its eigenvalues and corresponding algebraic multiplicities; because:
$$\begin{vmatrix}
    2-\lambda & 1 & 0 \\
    0 & 0-\lambda & 2 \\
    0 & 3 & -1-\lambda \\
    \end{vmatrix}=\left(2-\lambda\right)\begin{vmatrix}
     0-\lambda & 2 \\
     3 & -1-\lambda \\
    \end{vmatrix}=\ldots$$
