HMMT February

By the factor theorem, $f(x)=a(x-u)(x-v)$, so the constraints essentially boil down to $2=f(w)=a(w-u)(w-v)$. We want to maximize the discriminant $b^{2}-4 a c=a^{2}\left[(u+v)^{2}-4 u v\right]=a^{2}(u-v)^{2}=a^{2}[(w-v)-(w-u)]^{2}$. 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 $(-2)^{2} \cdot 2^{2}=16$.