AMC10 2025 A

AMC10 2025 A · Q7

AMC10 2025 A · Q7. It mainly tests Linear equations, Systems of equations.

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 $b-a$?

设$a$和$b$是实数。当多项式$x^3+x^2+ax+b$被$x-1$除时，余数是$4$。当被$x-2$除时，余数是$6$。$b-a$是多少？

(A) 14 14

(B) 15 15

(D) 17 17

(E) 18 18

Answer

Correct choice: (E)

正确答案：（E）

Solution

Use synthetic division to find that the remainder of $x^{3}+x^{2}+ax+b$ 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)=18

使用合成除法发现$x^{3}+x^{2}+ax+b$被$x-1$除时的余数是$a+b+2$，被$x-2$除时的余数是$2a+b+12$。现在，我们解 \[\begin{cases} a+b+2=4 \\ 2a+b+12 = 6 \\ \end{cases}\] 最终得到$a=-8$，$b=10$，所以$b-a=10-(-8)=18$

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