If $A$ is singular, is $A^3+A^2+A$ singular?
Since $A$ is singular, it has a non-trivial kernel. Let $v$ be a non-zero vector killed by $A$.
Show that $A^3+A^2+A$ kills $v$ too.
$A$ is singular hence $|A|=0$ thus: $$|A^{3}+A^{2}+A|=|A(A^{2}+A+I)|=|A||A^{2}+A+I|=0\cdot|A^{2}+A+I|=0$$
hence $A^{3}+A^{2}+A$ is also singular
If $A$ is singular, then $AB$ is singular too, for any choice of $B$. Just put $B=A^2+A+I$.