AMC12 2001 A

AMC12 2001 A · Q23

AMC12 2001 A · Q23. It mainly tests Polynomials, Complex numbers (rare).

A polynomial of degree four with leading coefficient 1 and integer coefficients has two zeros, both of which are integers. Which of the following can also be a zero of the polynomial?

一个四次多项式的首项系数为 1，且系数均为整数。它有两个零点，并且这两个零点都是整数。下列哪一个也可能是该多项式的零点？

(A) $\frac{1+i\sqrt{11}}{2}$ $\frac{1+i\sqrt{11}}{2}$

(B) $\frac{1+i}{2}$ $\frac{1+i}{2}$

(D) $1 + \frac{i}{2}$ $1 + \frac{i}{2}$

(E) $\frac{1+i\sqrt{13}}{2}$ $\frac{1+i\sqrt{13}}{2}$

Answer

Correct choice: (A)

正确答案：（A）

Solution

Let the polynomial be $P$ and let the two integer zeros be $z_1$ and $z_2$. We can then write $P(x)=(x-z_1)(x-z_2)(x^2+ax+b)$ for some integers $a$ and $b$. If a complex number $p+qi$ with $q\not=0$ is a root of $P$, it must be the root of $x^2+ax+b$, and the other root of $x^2+ax+b$ must be $p-qi$. We can then write $x^2+ax+b = (x-p-qi)(x-p+qi) = (x-p)^2 - (qi)^2 = x^2 - 2px + p^2 + q^2$. We can now examine each of the five given complex numbers, and find the one for which the values $-2p$ and $p^2+q^2$ are integers. This is $\boxed{\frac {1 + i \sqrt {11}}{2}}$, for which we have $-2p = -2\cdot\frac 12 = -1$ and $p^2+q^2 = \left( \frac 12 \right)^2 + \left( \frac {\sqrt{11}}2 \right)^2 = \frac 14 + \frac {11}4 = \frac {12}4 = 3$. (As an example, the polynomial $x^4 - 2x^3 + 4x^2 - 3x$ has zeroes $0$, $1$, and $\frac {1 \pm i \sqrt {11}}{2}$.)

设该多项式为 $P$，两个整数零点为 $z_1$ 和 $z_2$。则可写成 $P(x)=(x-z_1)(x-z_2)(x^2+ax+b)$，其中 $a$ 与 $b$ 为整数。若复数 $p+qi$（其中 $q\not=0$）是 $P$ 的一个根，则它必为 $x^2+ax+b$ 的根，并且 $x^2+ax+b$ 的另一个根必须是 $p-qi$。于是 $x^2+ax+b = (x-p-qi)(x-p+qi) = (x-p)^2 - (qi)^2 = x^2 - 2px + p^2 + q^2$。现在检查给出的五个复数，找出使得 $-2p$ 与 $p^2+q^2$ 都为整数的那个。答案是 $\boxed{\frac {1 + i \sqrt {11}}{2}}$，因为此时 $-2p = -2\cdot\frac 12 = -1$，且 $p^2+q^2 = \left( \frac 12 \right)^2 + \left( \frac {\sqrt{11}}2 \right)^2 = \frac 14 + \frac {11}4 = \frac {12}4 = 3$。（例如，多项式 $x^4 - 2x^3 + 4x^2 - 3x$ 的零点为 $0$、$1$、以及 $\frac {1 \pm i \sqrt {11}}{2}$。）

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