HMMT February

Solution:
Answer: $16$

By the factor theorem, $f(x) = a(x-u)(x-v)$, so the constraints essentially boil down to $2 = f(w) = a(w-u)(w-v)$. (It's not so important that $u \neq v$; we merely specified it for a shorter problem statement.)

We want to maximize the discriminant $b^{2} - 4 a c = a^{2}[(u+v)^{2} - 4 u v] = a^{2}(u-v)^{2} = a^{2}[(w-v)-(w-u)]^{2}$. Clearly $a \mid 2$. If $a > 0$, then $(w-u)(w-v) = 2 / a > 0$ means the difference $|u-v|$ is less than $2 / a$, whereas if $a < 0$, since at least one of $|w-u|$ and $|w-v|$ equals $1$, the difference $|u-v|$ of factors is greater than $2 / |a|$.

So the optimal choice occurs either for $a = -1$ and $|u-v| = 3$, or $a = -2$ and $|u-v| = 2$. The latter wins, giving a discriminant of $(-2)^{2} \cdot 2^{2} = 16$.