Question

设 $P(x)$ 是有理系数多项式，满足：用 $x^2 + x + 1$ 除 $P(x)$ 的余式为 $x + 2$ ，用 $x^2 + 1$ 除 $P(x)$ 的余式为 $2x + 1$ 。满足这两个性质的最低次数多项式唯一。该多项式各系数平方和是多少？

Let $P(x)$ be a polynomial with rational coefficients such that when $P(x)$ is divided by the polynomial $x^2 + x + 1,$ the remainder is $x + 2,$ and when $P(x)$ is divided by the polynomial $x^2 + 1,$ the remainder is $2x + 1.$ There is a unique polynomial of least degree with these two properties. What is the sum of the squares of the coefficients of that polynomial?

$10$

$13$

$19$

$20$

$23$

答案：E

知识点：多项式中国剩余定理

难度评级：2020

小提示：

最低次数多项式是三次多项式；将它写成 $P(x) = x + 2$ $+ (x^2 + x + 1)(px + q)$ 。

The least-degree polynomial is a cubic; write $P(x) = x + 2$ $+ (x^2 + x + 1)(px + q)$

大提示：

模 $x^2 + 1$ 化简这个式子，利用 $x^2 \equiv -1$ ，再将所得余式与 $2x + 1$ 比较。

Reduce that expression modulo $x^2 + 1$ using $x^2 \equiv -1,$ and match it to $2x + 1$

解答：

一次多项式不可能给出这两个不同的余式。若 $P(x)=ax^2+bx+c$ 是二次的，比较它的两个余式会同时得到 $b-a=1,\ c-a=2$ 和 $b=2,\ c-a=1$ ，互相矛盾。因此最低可能次数是 $3$ 。写成 $P(x) = (x + 2)$ $+ (x^2 + x + 1)(px + q)$ ，它除以 $x^2 + x + 1$ 的余式就是 $x + 2$ 。

模 $x^2 + 1$ 化简（即令 $x^2 \equiv -1$ ），所得余式为 $(q + 1)x + (2 - p)$ 。令它等于 $2x + 1$ ，得到 $q = 1$ 且 $p = 1$ 。

于是 $P(x) = x^3 + 2x^2 + 3x + 3$ ，各系数的平方和为 $1 + 4 + 9 + 9 = 23$ 。

所以正确答案是 E。

No linear polynomial can have the two different remainders. If $P(x)=ax^2+bx+c$ were quadratic, comparing its remainders would give both $b-a=1,\ c-a=2$ and $b=2,\ c-a=1,$ a contradiction. Thus the least possible degree is $3.$ Write $P(x) = (x + 2)$ $+ (x^2 + x + 1)(px + q),$ which has remainder $x + 2$ upon division by $x^2 + x + 1.$

Reducing modulo $x^2 + 1$ (so $x^2 \equiv -1$ ) gives remainder $(q + 1)x + (2 - p).$ Setting this equal to $2x + 1$ gives $q = 1$ and $p = 1.$

Then $P(x) = x^3 + 2x^2 + 3x + 3,$ and the sum of the squares of the coefficients is $1 + 4 + 9 + 9 = 23.$

Thus, the correct answer is E.

2022 AMC 12B 第 20 题

答案：E

难度评级：2020

小提示：

大提示：

解答：

其他年份的第 20 题