2017 AMC 12A Problem 23

Since $g$ has three distinct roots all shared by the quartic $f,$ we can write $f(x)=(x-q)g(x)$ for some remaining root $q.$ Expanding, $$\begin{aligned} &f(x)=x^4+(a-q)x^3 \\ &\quad {}+(1-qa)x^2 \\ &\quad {}+(10-q)x-10q. \end{aligned}$$ Matching the $x$ coefficient, $10-q=100,$ so $q=-90.$ Matching the $x^3$ coefficient, $a-q=1,$ so $a=-89.$ Then $g(1)=1+a+1+10$ $=12-89=-77$ and $1-q=91,$ so $$\begin{aligned} &f(1)=(1-q)g(1) \\ &=91\cdot(-77)=-7007. \end{aligned}$$ Thus, the correct answer is C.