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2025 AMC 10A Problems/Problem 7

Suppose $a$ and $b$ are real numbers. When the polynomial $x^3+x^2+ax+b$ is divided by $x-1$, the remainder is $4$. When the polynomial is divided by $x-2$, the remainder is $6$. What is $b-a$?

$\textbf{(A) } 14 \qquad \textbf{(B) } 15 \qquad \textbf{(C) } 16 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 18$

Solution 1

Use synthetic division to find that the remainder when $x^{3}+x^{2}+ax+b$ is $a+b+2$ when divided by $x-1$ and $2a+b+12$ when divided by $x-2$. Now, we solve

\[\begin{cases} a+b+2=4 \\ 2a+b+12 = 6 \\ \end{cases}\]

This ends up being $a=-8$, $b=10$, so $b-a=10-(-8)=\fbox{\textbf{(E)} 18}$

Solution 2

Via the remainder theorem, we can plug $1$ in for the factor $x-1$ and get $4$, so we have that \[1^3+1^2+1a+b=4\] \[a+b=2.\] Then, we plug in $2$ and get a remainder of $6$, so we have that \[2^3+2^2+2a+b=6\] \[8+4+2a+b=6\] \[2a+b=-6\] Then, we solve the system of equations. By substitution, we obtain \[b=2-a\] \[2a+2-a=-6\] \[a=-8\] \[b=2-(-8)=2+8=10.\] Thus, we have that $b-a=10-(-8)=10+8=\fbox{\textbf{(E)} 18}$

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