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

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

解答：

其他年份的第 19 题