Question 730·Hard·Math

No Desmos needed. The $x^3$ term gives the sum of four roots directly. An $x$-intercept has $y=0$, so the four listed inputs make $g$ equal $0$: $g(a)=g(b)=g(c)=g(d)=0$. These are all four roots, or inputs giving output $0$: a fourth-degree polynomial can't have more than four distinct roots.

The problem defines $g(x)=f(x)+60$. Since $f(x)$ has no term without $x$, adding $60$ changes only the constant term:

Because the coefficient of $x^4$ is $1$, the four roots let you write $g(x)=(x-a)(x-b)(x-c)(x-d)$. To get an $x^3$ term, take $-a$, $-b$, $-c$, or $-d$ from one factor and $x$ from the other three. Their contributions add to $-(a+b+c+d)x^3$. The $x^3$ coefficient is the negative of the sum of the four roots. That's the sum-of-roots rule here.

The $x^3$ coefficient in $g(x)$ is $-14$, so match it to the expression from the factors: $-(a+b+c+d)=-14$. Multiply by $-1$: $a+b+c+d=14$. The sum of the four $x$-coordinates is $14$. Choice B.