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

Revision as of 19:34, 6 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 $b-a$ ?

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

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

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) $6 = (2)^3 + (2)^2 + (2)a + b = 8 + 4 + 2a + b$ so we get $-6 = 2a + b$ (ii)

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

So $b - a = 10 - (-8) = \boxed{18}$

Video Solution (In 1 Min)

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

Video Solution

https://youtu.be/gWSZeCKrOfU

~MK

@@ Line 37: / Line 37: @@
 <imath>6 = (2)^3 + (2)^2 + (2)a + b = 8 + 4 + 2a + b</imath> so we get <imath>-6 = 2a + b</imath> (ii)
-So we Subtract (ii) from (i) to get a = -8. Substitute in to get b = 10.
+So we Subtract (ii) from (i) to get <imath>a = -8</imath>. Substitute in to get <imath>b = 10</imath>.
-So $b - a = 10 - (-8) = \boxed{18(\text{E})}
+So <imath>b - a = 10 - (-8) = \boxed{18}</imath>
 == Video Solution (In 1 Min) ==

2025 AMC 10A (Problems • Answer Key • Resources)
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