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2025 AMC 10A Problems/Problem 7: Difference between revisions

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So we Subtract (ii) from (i) to get <imath>a = -8</imath>. Substitute in to get <imath>b = 10</imath>.
So we Subtract (ii) from (i) to get <imath>a = -8</imath>. Substitute in to get <imath>b = 10</imath>.


So <imath>b - a = 10 - (-8) = \boxed{\textbf{(E)}18}</imath>
So <imath>b - a = 10 - (-8) = \fbox{\textbf{(E)} 18}</imath>


~Aarav22
~Aarav22

Revision as of 20:49, 7 November 2025

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 $b-a$?

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

Video Solution

https://youtu.be/l1RY_C20Q2M

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 $b-a=10-(-8)=10+8=\fbox{\textbf{(E)} 18}$

~ranu540

Solution 3 (Small)

Using Remainder theorem, we get that: $4 = (1)^3 + (1)^2 + (1)a + b = 1 + 1 + a + b$ so we get $2 = a + b$ (i) and from the second statement, $6 = (2)^3 + (2)^2 + (2)a + b = 8 + 4 + 2a + b$ so we get $-6 = 2a + b$ (ii).

So we Subtract (ii) from (i) to get $a = -8$. Substitute in to get $b = 10$.

So $b - a = 10 - (-8) = \fbox{\textbf{(E)} 18}$

~Aarav22

Chinese Video Solution

https://www.bilibili.com/video/BV1t72uBREvf/

~metrixgo

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=dAeyV60Hu5c

Video Solution (In 1 Min)

https://youtu.be/pDk05d9r-4c?si=PIZ1YHPfXFoEKUKW ~ Pi Academy

Video Solution

https://youtu.be/gWSZeCKrOfU

~MK

Video Solution by Daily Dose of Math

https://youtu.be/gPh9w3X3QSw

~Thesmartgreekmathdude

See Also

2025 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America.

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