Determine for with values of $a$ the matrix is diagonalizable over $\mathbb{R}$ 
Determine for which values of $a$ \begin{pmatrix} 4 & 0 & 0 \\ 4 & 4 & a \\ 4 & 4 & 4 \end{pmatrix}
The matrix is diagonalizable

So we first look at the characteristic polynomial:
$$ \begin{vmatrix} 4-\lambda & 0 & 0 \\ 4 & 4-\lambda & a \\ 4 & 4 & 4-\lambda \end{vmatrix}=(4-\lambda)\begin{vmatrix}  4-\lambda & a \\  4 & 4-\lambda \end{vmatrix}=\\=(4-\lambda)[(4-\lambda)^2-4a]=(4-\lambda)[(\lambda^2-8\lambda+16-4a]=\\=4\lambda^2-32\lambda+64-16a-\lambda^3+8\lambda^2-16\lambda+4\lambda a=\lambda^3+12\lambda^2-48+64-16a+4\lambda a$$
How can I find the roots to this polynomial? usually with 3rd power polynomial I use to find the factor of the free element and test to find when the polynomial$=0$
What Should I do in this case?
 A: You can do this without much computation.
First off, for $a=0$ the matrix is triangular (but not diagonal) with equal diagonal entries, therefore not diagonalisable.
Then assume $a\neq 0$. Now the matrix is block triangular, so its characteristic polynomial is the product of those of the diagonal blocks $D_1=(4)$ and $D_2=(\begin{smallmatrix}4&a\\4&4\end{smallmatrix})$. The former characteristic polynomial is $X-4$ with root $4$, while the characteristic polynomial of $D_2$ is a quadratic polynomial with the sum of its roots being (the trace) $8$. Now $4$ is clearly not a root of the latter characteristic polynomial (as $D_2-4I$ is invertible since $a\neq0$), an consequently $D_2$ cannot have multiple roots over $\Bbb C$, so the matrix is diagonalisable over $\Bbb C$.
Finally, since the question was about being diagonalisable over$~\Bbb R$, one does have to know the discriminant of the characteristic polynomial of $D_2$. Here I do have to admit some computation: the discriminant is $64-4\times\det D_2 =16 a$, so the eigenvalues of $D_2$ are real (and the original matrix diagonalisable over$~\Bbb R$) if and only if $a>0$.
