Maybe off topic. For polynomial:
$${x}^{5}+q\,{x}^{4}+p\,{x}^{3}+t\,{x}^{2}+k\,x-m=0$$
with root R, characteristic polinomial of matrix:
$$\small\begin{pmatrix}a & -c\,t-b\,p+g\,m-d\,k & -b\,t+d\,m-c\,k & c\,m-b\,k & b\,m\cr b & b\,q+a & -c\,t+g\,m-d\,k & d\,m-c\,k & c\,m\cr c & c\,q+b & b\,q+c\,p+a & g\,m-d\,k & d\,m\cr d & d\,q+c & c\,q+d\,p+b & d\,t+b\,q+c\,p+a & g\,m\cr g & g\,q+d & d\,q+g\,p+c & g\,t+c\,q+d\,p+b & d\,t+b\,q+c\,p+g\,k+a\end{pmatrix}$$
can have a root:
$$\small{x=g\,{R}^{4}+\left( g\,q+d\right) \,{R}^{3}+\left( d\,q+g\,p+c\right) \,{R}^{2}+\left( g\,t+c\,q+d\,p+b\right) \,R+d\,t+b\,q+c\,p+g\,k+a}$$
For example:$${x}^{5}-m=0$$
characteristic polinomial of matrix:
$$\begin{pmatrix}a & g\,m & d\,m & c\,m & b\,m\cr b & a & g\,m & d\,m & c\,m\cr c & b & a & g\,m & d\,m\cr d & c & b & a & g\,m\cr g & d & c & b & a\end{pmatrix}$$
has a root:
$$x=g\,{m}^{\frac{4}{5}}+d\,{m}^{\frac{3}{5}}+c\,{m}^{\frac{2}{5}}+b\,{m}^{\frac{1}{5}}+a$$
For example:
$${x}^{5}+{x}^{4}-4\,{x}^{3}-3\,{x}^{2}+3\,x+1=0$$
has a root:
$$x=2\,\mathrm{cos}\left( \frac{2\,\pi \,k}{11}\right)$$
if a=1,b=1,c=1,d=1,g=1
characteristic polinomial
$$-{x}^{5}-6\,{x}^{4}+{x}^{3}+10\,{x}^{2}+6\,x+1$$
has a root:
$$x=16\,{\mathrm{cos}\left( \frac{2\,\pi \,k}{11}\right) }^{4}+16\,{\mathrm{cos}\left( \frac{2\,\pi \,k}{11}\right) }^{3}-8\,{\mathrm{cos}\left( \frac{2\,\pi \,k}{11}\right) }^{2}-10\,\mathrm{cos}\left( \frac{2\,\pi \,k}{11}\right) -2$$
Sorry for my bad English.
Best Answer
The roots of $f(x)$ are $$x=-4,-2,2,4$$, according to your graph, so for $f(f(x)) = 0$, we have $$f(x) = -4,-2,2,4$$ because $f(x)$ took the place of $x$. So, we need to examine when $f(x) = -4,-2,2,4$.
From the graph we see that $f(x)$ is never $-4$
From the graph we see that $f(x)=-2$ at $2$ values of $x$.
From the graph we see that $f(x) = 2$ at $4$ values of $x$
From the graph we see that $f(x) = 4$ at $2$ values of $x$
Thus, that makes $8$ values of $x$ total that $f(f(x)) = 0 $, so it has $8$ roots.