TITLE: Let $f(x)=x^2+12x+30$. Solve $f(f(f(f(f(x)))))=0$
QUESTION [21 upvotes]: Here is my solve, is it correct?
I figured out that we can restate $f(f(x))$ as 
$((x+r)(x+s)+r)((x+r)(x+s)+s)$ 
thus
 $f(f(f(f(f(x)))))=0$ 
is 
$(x+r)^2(x+s)^2(4s+3r)(4r+3s)$
from vieta's
$0=(x+r)(x+s)(36^2-r^2)$ and we get $x=-6$
Did I make a mistake somewhere?
I'd also like to see other solutions
EDIT: How can I solve it using vieta then?

REPLY [3 votes]: Define inductively $f^n(x)=f(f^{n-1}(x))$ for all $n\geq1$, so that you are solving $f^5(x)=0$.
We have $f^5(x)=f(f^4(x))$ thus $f^4(x)$ is either solution of $x^2+12x+30=0$, i.e.
$$
f^4(x)=-3\pm\sqrt{6}.
$$
Then for the same reason $f^3(x)$ is any value $z$ such that
$$
z^2+12z+30=-3\pm\sqrt{6}.
$$
Iterating this process a few more times you get to the solutions of $f^5(x)=0$.