2001 AMC 12 第 19 题

Question

多项式

P(x) = x^3 + ax^2 + bx + c

具有如下性质：它的零点的平均数、零点的乘积以及各项系数之和都相等。若函数

y = P(x)

的图像的

y

-截距是

2

，则

b

是多少？

The polynomial

P(x) = x^3 + ax^2 + bx + c

has the property that the mean of its zeros, the product of its zeros, and the sum of its coefficients are all equal. If the

y

-intercept of the graph of

y = P(x)

is

2,

what is

b?

-11

-10

-9

1

5

答案：A

知识点：韦达定理多项式

难度评级：1760

小提示：

由韦达定理，零点之和是

-a

，零点之积是

-c

；系数之和是

P(1)

By Vieta, the sum of the zeros is

-a

and the product is

-c;

the sum of coefficients is

P(1)

大提示：

y

-截距给出

c = 2

，所以公共值就是零点的乘积

The

y

-intercept gives

c = 2,

so the common value is the product of the zeros

解答：

y

-截距为

P(0) = c = 2

。由韦达定理，零点之积为

-c = -2

，零点的平均数为

-\dfrac{a}{3}

，系数之和为

P(1) = 1 + a + b + c

。

这三者都等于

-2

。由

-\dfrac{a}{3} = -2

得

a = 6

。

于是

1 + a + b + c = -2

变为

1 + 6 + b + 2 = -2

，所以

b = -11

。

因此，正确答案是 A。

The

y

-intercept is

P(0) = c = 2.

By Vieta’s formulas the product of the zeros is

-c = -2,

the mean of the zeros is

-\dfrac{a}{3},

and the sum of the coefficients is

P(1) = 1 + a + b + c.

All three are equal to

-2.

From

-\dfrac{a}{3} = -2

we get

a = 6.

Then

1 + a + b + c = -2

becomes

1 + 6 + b + 2 = -2,

so

b = -11.

Thus, the correct answer is A.

2001 AMC 12 第 19 题

答案：A

难度评级：1760

小提示：

大提示：

解答：

其他年份的第 19 题