Find an invertible matrix V and a diagonal matrix D so that A=VDV^-1
Either you have written the matrix incorrectly or you have the wrong eigenvalues. The "characteristic equation" is
does NOT satisfy that. In fact, the eigenvalues are . Since there are two distinct eigenvalues, there are two independent eigenvectors and the matrix is "diagonalizable" with the and V equal to the matrix having the eigenvectors as columns.
On the other hand, if either the "2" or "8" in the matrix is negative, then the characteristic equation is and has is a "double" eigenvalue. Now you need to find the eigenvectors. If there are two independent eigenvectors corresponding to eigenvalue 1, then A is "diagonalizable": with D being the diagonal matrix , with the eignvalues along the diagonal, and V is the matrix having two independent eigenvectors as columns. If there are not two independent eigenvalues (all eigenvalues can be written as a multiple of one eigenvector) then the matrix is not diagonalizable.
(But, in that case, it could be written in "Jordan Normal Form" .)