2017 AMC 10A Problem 24

Since the three distinct roots of $g$ are roots of $f$, the cubic $g(x)$ divides $f(x)$. Both are monic, so the quotient is monic and linear:
$$f(x) = g(x)(x - r) = (x^3 + ax^2 + x + 10)(x - r)$$
for some real $r$.

Expand and match coefficients with $x^4 + x^3 + bx^2 + 100x + c$:
- $x^3$: $a - r = 1$.
- $x$: $10 - r = 100$, so $r = -90$.
- $x^2$: $1 - ar = b$; constant: $-10r = c$.

From $r = -90$ we get $a = 1 + r = -89$. We do not need $b$ or $c$, because
$$f(1) = g(1)\,(1 - r) = (1 + a + 1 + 10)(1 + 90) = (12 - 89)(91) = (-77)(91) = -7007.$$

The answer is $\boxed{\textbf{(C)}\ -7007}$.

Sharing all roots of a lower-degree polynomial means divisibility, and monic-ness fixes the quotient up to one unknown. Coefficient matching then needs only as many equations as unknowns; here the linear coefficient alone determines $r$. Evaluating the factored form at $x = 1$ avoids computing $b$ and $c$ at all.

Finish the expansion: $b = 1 - (-89)(-90) = 1 - 8010 = -8009$ and $c = -10(-90) = 900$. Then $f(1) = 1 + 1 - 8009 + 100 + 900 = -7007$, confirming the answer.

• Trying to find the roots of g, or solving for b and c by full expansion and making an arithmetic slip; f(1) = g(1)(1 - r) needs only a and r.
• Writing the linear factor as $(x + r)$ but then using the sign convention of $(x - r)$, which flips the sign of $r$ and gives $g(1) \cdot (-89)$ or similar.