HMMT-Nov 2022 guts, Problem 21

Let $P(x)$ be a quadratic polynomial with real coefficients. Suppose that $P(1)=20, P(-1)=22$, and $P(P(0))=400$. Compute the largest possible value of $P(10)$.

Solution: Solution: Let $P(x)=a x^{2}+b x+c$. The given equations give us: \[ \begin{aligned} & a+b+c=20 \\ & a-b+c=22 \end{aligned} \] Hence $b=-1, a+c=21$, and so the final equation gives us $a c^{2}=400$. Substituting $a=21-c$ and solving the cubic in $c$, we get $c=-4,5,20$. Of these, the smallest value $c=-4$ (and hence $\left.P(x)=25 x^{2}-x-4\right)$ ends up giving the largest value of $P(10)$. $\fbox{2486}$.

Answer: 2486