2022 AMC 12B 第 20 题

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：

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