TITLE: $f(g(h(x)))=0$ has $8$ real roots
QUESTION [10 upvotes]: Find all quadratic polynomials $f(x),g(x)$ and $h(x)$ such that the polynomial $f(g(h(x)))=0$ has roots $1,2,3,4,5,6,7$ and $8$.
I don't know what to do. Making a $8$ degree equation is quite tedious. Thanks.

REPLY [3 votes]: The polynomial $$F(x)=\prod_{1\le k\le 8}(x-k)$$ has a derivative equal to
$$4(2x-9)(x^6-27x^5+288x^4-1539x^3+4299x^2-5886x+3044)$$ where the second 6-degree factor is irreducible (verify this, for example, in WolframAlpha). 
Therefore if $F(x)=f(g(h(x)))$  then 
$$F’(x)=f’(g(h(x)))\cdot g’(h(x))\cdot h’(x)$$ which is a contradiction because $F’(x)$ is a product of only two irreducible factors.