Problem 11.6. The quadratic trinomial $P(x)$ is such that $P(P(x))=x^{4}-2 x^{3}+4 x^{2}-3 x+4$. What can $P(8)$ be? List all possible options.
Answer: 58.

Solution. Let $P(x)=a x^{2}+b x+c$. Then $P(P(x))=a\left(a x^{2}+b x+c\right)^{2}+b\left(a x^{2}+b x+c\right)+c=$
$a\left(a^{2} x^{4}+2 a b x^{3}+\left(b^{2}+2 a c\right) x^{2}+2 b c x+c^{2}\right)+b\left(a x^{2}+b x+c\right)+c$. Therefore,

\[
\begin{aligned}
& x^{4}-2 x^{3}+4 x^{2}-3 x+4=P(P(x))= \\
& =a^{3} x^{4}+2 a^{2} b x^{3}+\left(a b^{2}+2 a^{2} c+a b\right) x^{2}+\left(2 a b c+b^{2}\right) x+\left(a c^{2}+b c+c\right)
\end{aligned}
\]

Equating the corresponding coefficients, we get the system of equations

\[
\left\{\begin{array}{l}
a^{3}=1 \\
2 a^{2} b=-2 \\
a b^{2}+2 a^{2} c+a b=4 \\
2 a b c+b^{2}=-3 \\
a c^{2}+b c+c=4
\end{array}\right.
\]

From the first equation, it immediately follows that $a=1$. Substituting $a=1$ into the second equation, we get $2 b=-2$, hence $b=-1$. Substituting $a=1$ and $b=-1$ into the third equation, we get $1+2 c-1=4$, hence $c=2$. It is clear that $a=1, b=-1, c=2$ is a solution to this system, so $P(x)=x^{2}-x+2$ and $P(8)=8^{2}-8+2=58$.