Hint: Given a characteristic equation, how can you obtain the eigen-values? (Also if you have repeated eigen-values what does this mean for the roots of the characteristic polynomial?)
An n by n matrix is diagonalizable if and only it has n independent eigenvectors. It can be shown that "distinct" (i. e "different") eigenvalues have eigenvectors that are independent eigenvectors so if an n by n matrix has n distinct eigenvalues, it has n independent eigenvectors and so is diagonalizable. The converse is NOT true. If a matrix has multiple eigenvalues it may or may not be diagonalizable.
For example if
Then each of A, B, and C have "3" as a triple eigenvalue. A has three independent eigenvectors, B has two independent eigenvectors, and C has only one independent eigenvector.
A can be "diagonalize" (clearly it is diagonal) but B and C cannot.