What is the sum of all the roots of the equation $f^{[5]}(x)=0$? 
Suppose $f(x)=x^2-3x+2$, what is the sum of all the roots of the equation $f^{[5]}(x)=0$?

I think one of the common ways of solving questions about $f^{[n]}(x)$ is fixed-point iteration.
I used this method, solving $x^2-3x+2=x$ and got $x=2 \pm \sqrt{2}$.
Therefore $$f(x)-(2+\sqrt{2})=[x-(2+\sqrt{2})][x+(\sqrt{2}-1)]......(1)$$
And $$f(x)-(2-\sqrt{2})=[x-(2-\sqrt{2})][x-(\sqrt{2}+1)]......(2)$$
I tried $\frac{(1)}{(2)}$ and immediately got trapped since I can’t see the next step. The reason why I wanted to know what $f^{[5]}(x)$ is is that if I can know the coefficient of $x^{31}$, I can easily obtain the answer by Vieta’s theorem.
Maybe I’m missing something obvious here. Any suggestions or hints would be appreciated.
Edit: I just realized different regions may have different expressions on this one.
Basically, $f^{[2]}(x)=f(f(x))$, $f^{[3]}(x)=f(f(f(x)))$, and so on. I am really sorry for not considering this trouble in the first place.

It seems obvious that, as you apply $f$, the degree of the polynomial doubles. (Proof: induction.)
Let $f^{[5]}(x)=x^{32}+ax^{31}+\ldots$.
As $f^{[4]}(x)=x^{16}+bx^{15}+\ldots$, this means that $f^{[5]}(x)=(x^{16}+bx^{15}+\ldots)^2-3(x^{16}+bx^{15}+\ldots)+2=x^{32}+2bx^{31}+\ldots$, so $a=2b$.
Similarly, as $f^{[3]}(x)=x^8+cx^7+\ldots$, this means that $f^{[4]}(x)=(x^8+cx^7+\ldots)^2-3(x^8+cx^7+\ldots)+2=x^{16}+2cx^{15}+\ldots$, so $b=2c$.
Again, $f^{[2]}(x)=x^4+dx^3+\ldots$, so as $f^{[3]}(x)=(x^4+dx^3+\ldots)^2-3(x^4+dx^3+\ldots)+2=x^8+2dx^7+\ldots$, we have $c=2d$.
Finally, $f(x)=x^2-3x+2$, so $f^{[2]}(x)=(x^2-3x+2)^2-3(x^2-3x+2)+2=x^4-6x^3+\ldots$ implies $d=-6$
Altogether, we have $d=-6, c=-12, b=-24, a=-48$. So, the sum of all zeros is $48$.
