2025 AMC 10A Problem 7

Write $P(x) = x^3 + x^2 + ax + b$. Dividing by $x - c$ leaves remainder $P(c)$, so the two conditions are $P(1) = 4$ and $P(2) = 6$:
$$P(1) = 1 + 1 + a + b = 4 \implies a + b = 2,$$
$$P(2) = 8 + 4 + 2a + b = 6 \implies 2a + b = -6.$$
Subtracting the first from the second gives $a = -8$, and then $b = 2 - a = 10$. Hence
$$b - a = 10 - (-8) = 18.$$
As a check, $P(x) = x^3 + x^2 - 8x + 10$ gives $P(1) = 4$ and $P(2) = 8 + 4 - 16 + 10 = 6$.

The answer is $\boxed{\textbf{(E)}\ 18}$.

The Remainder Theorem replaces a division by a single evaluation, and since $a$ and $b$ enter $P$ linearly, each evaluation is a linear equation in them. Two conditions therefore pin down two unknowns with no polynomial division at all. The same idea scales: $k$ remainder conditions at $k$ distinct points determine $k$ unknown coefficients.

Subtract before solving. The fixed part of $P$ contributes $(8 - 1) + (4 - 1) = 10$ to $P(2) - P(1)$, and the $ax$ term contributes $a$, while $b$ cancels:
$$P(2) - P(1) = 10 + a = 6 - 4 = 2 \implies a = -8,$$
and $P(1) = 4$ then gives $b = 10$.

• Substituting the divisors' coefficients rather than their roots, that is evaluating at x = -1 and x = -2, which yields b - a = 4 and no listed choice.
• Solving correctly and then reporting $a - b = -18$, or reporting $b = 10$ alone.
• Dropping the $x^2$ term when evaluating at $x = 2$, which turns $2a + b = -6$ into $2a + b = -2$.