I have a $3\times 3$ Matrix polynomial which I computed the characteristic polynomial to be $x^3 - 2x^2 + x = x(x-1)^2$. This would give us Eigenvalues 0,1.
Seeing how there are 2 eigenvalues and $n = 3$ for this matrix. This would usually mean this matrix cannot be diagonalizable. However, I'm wondering if the multiplicity of 2 affects our eigenvalues?
It may or may not be diagonalizable.
An example of such a matrix that is diagonalizable is $$\left(\begin{array}{ccc} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 0 \end{array}\right).$$
An example of such a matrix that is not diagonalizable is $$\left(\begin{array}{ccc} 1 & 1 & 0\\ 0 & 1 & 0\\ 0 & 0 & 0 \end{array}\right).$$
(I'll leave it to you to establish that the first matrix is diagonalizable and the second one is not)
In short, the information you have is insufficient to determine whether the matrix is diagonalizable or not.
From comments, the matrix in question is $$A=\left(\begin{array}{rrr} -2& 2 & 1\\ -4&4 & 2\\ 1 & -1 & 0 \end{array}\right).$$ If you check the nullity of $A-I$ (what you need to find the eigenvectors corresponding to $1$), you will find that the rank of $A-I$ is $2$ and the nullity is $1$. That means the geometric multiplicity of $\lambda=1$ is $1$, strictly less than its algebraic multiplicity, so the matrix in question is not diagonalizable.