[Algebra]
Problem: The quadratic polynomial $f(x)$ has a zero at $x=2$. The polynomial $f(f(x))$ has only one real zero, at $x=5$. Compute $f(0)$.

Solution: Let $f(x)=a(x-b)^{2}+c$. The graph of $f$ is symmetric about $x=b$, so the graph of $y=f(f(x))$ is also symmetric about $x=b$. If $b \neq 5$, then $2 b-5$, the reflection of 5 across $b$, must be a zero of $f(f(x))$. Because $f(f(x))$ has exactly one zero, $b=5$.

Because $f(2)=0$ and $f$ is symmetric about $x=5$, the other zero of $f$ is $x=8$. Because the zeros of $f$ are at 2 and 8 and $f(5)$ is a zero of $f$, either $f(5)=2$ or $f(5)=8$. The following argument shows that $f(5)=8$ is impossible. Because $f$ is continuous, if $f(5)=8$, then $f\left(x_{0}\right)=2$ for some $x_{0}$ in the interval $2<x_{0}<5$. In that case, $f\left(f\left(x_{0}\right)\right)=0$, so 5 would not be a unique zero of $f(f(x))$. Therefore $f(5)=2$ and $c=2$. Setting $f(2)=0$ yields the equation $a(2-5)^{2}+2=0$, so $a=-\frac{2}{9}$, and $f(0)=-\frac{\mathbf{3 2}}{\mathbf{9}}$.

Final answer: $-\frac{32}{9}$