2017 AMC 10A Problem 24

1. Why $f$ is $g$ times something. Call the roots of $g$ by $r_1, r_2, r_3$. Since $g$ starts with $x^3$, it factors as $g(x) = (x - r_1)(x - r_2)(x - r_3)$. These three roots are also roots of $f$, which starts with $x^4$ and so has one more root, call it $r$. So
$$f(x) = (x - r_1)(x - r_2)(x - r_3)(x - r) = g(x)\,(x - r).$$
2. Multiply it out, keeping track of each power of $x$:
$$(x^3 + ax^2 + x + 10)(x - r) = x^4 + (a - r)x^3 + (1 - ar)x^2 + (10 - r)x - 10r.$$
3. Match with $f(x) = x^4 + x^3 + bx^2 + 100x + c$, one power at a time. Start with the easiest line:
- $x$ terms: $10 - r = 100$, so $r = -90$.
- $x^3$ terms: $a - r = 1$, so $a = 1 + r = -89$.
4. Find $f(1)$ without $b$ or $c$. Put $x = 1$ into the factored form $f(x) = g(x)(x - r)$:
$$g(1) = 1 + a + 1 + 10 = 12 + (-89) = -77,$$
$$1 - r = 1 + 90 = 91,$$
$$f(1) = (-77)(91) = -7007.$$

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

If every root of one polynomial is also a root of a bigger one, the bigger one is the smaller one times an extra factor. Writing that out and matching the numbers in front of each power of $x$ turns the problem into a few small equations. Pick the easiest equation first; here the $x$ term gave $r$ right away.

Finish the matching and plug in. The $x^2$ terms give $b = 1 - ar = 1 - (-89)(-90) = 1 - 8010 = -8009$. The constant gives $c = -10r = 900$. Then
$$f(1) = 1 + 1 + (-8009) + 100 + 900 = -7007.$$

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