MHB Solving 8 Roots: Can 3 Quadratic Polynomials Fulfill $f(g(h(x)))=0$?

[anemone]

Is it possible to find three quadratic polynomials $f(x),\,g(x)$ and $h(x)$ such that the equation $f(g(h(x)))=0$ has the eight roots 1, 2, 3, 4, 5, 6, 7 and 8?

 

[anemone]

Spoiler: Solution of other

Suppose there are such $f,\,g,\,h$. Then $h(1),\,h(2),\cdots,\,h(8)$ will be the roots of the 4th degree polynomial $f(g(x))$. Since $h(a)=h(b),\,a\ne b$ if and only if $a$ and $b$ are symmetric with respect to the axis of the parabola, it follows that $h(1)=h(8),\,h(2)=h(7),\,h(3)=h(6),\,h(4)=h(5)$ and the parabola $y=h(x)$ is symmetric with respect to $x=\dfrac{9}{2}$. Also, we have either $h(1)<h(2)<h(3)<h(4)$ or $h(1)>h(2)>h(3)>h(4)$.

Now, $g(h(1)),\,g(h(2)),\,g(h(3)),\,g(h(4))$ are the roots of the quadratic polynomial $f(x)$, so $g(h(1))=g(h(4))$ and $g(h(2))=g(h(3))$ , which implies $h(1)+h(4)=h(2)+h(3)$. For $h(x)=Ax^2+Bx+C$, this would force $A=0$, a contradiction.