An $n\times n$ matrix $E$ has determinant $0$ and $E = D^2-8D+15I$, for some $n\times n$ matrix $D$. Argue that either $3,5$ or both is an eigenvalue of $D$.
I'm really struggling with this question. I know that $0$ is an eigenvalue of $E$, but that's about all I know. If someone could at least point me in the right direction as to how to think about this it would be appreciated.
Hint
$1)$ Solve the equation: $x^2-8x+15=0$ and see that $3$ and $5$ are roots, then you can write it as $(x-3)(x-5)=0$. It means that you can also write $D^2-8D+15I$ as $E=(D-3I)(D-5I)$
$2)$ Now use the known result $\det(AB)=\det A\det B$ at $\det E=\det[(D-3I)(D-5I)]$.
Can you finish?