Question

A polynomial $P(x) = c_{2004} x^{2004} + c_{2003} x^{2003} + \cdots + c_1 x + c_0$ has real coefficients with $c_{2004} \ne 0$ and $2004$ distinct complex zeros $z_k = a_k + b_k i,$ $1 \le k \le 2004$ with $a_k$ and $b_k$ real, $a_1 = b_1 = 0,$ and $\sum_{k=1}^{2004} a_k = \sum_{k=1}^{2004} b_k.$ Which of the following quantities can be a nonzero number?

$c_0$

$c_{2003}$

$b_2 b_3 \ldots b_{2004}$

$\displaystyle\sum_{k=1}^{2004} a_k$

$\displaystyle\sum_{k=1}^{2004} c_k$

Solution:

Since $z_1 = a_1 + b_1 i = 0$ is a root, $c_0 = P(0) = 0.$

The nonreal zeros occur in conjugate pairs, so $\sum b_k = 0,$ and the hypothesis then forces $\sum a_k = 0.$ The coefficient $c_{2003}$ equals $-c_{2004}$ times the sum of the roots $\sum a_k + i \sum b_k = 0,$ so $c_{2003} = 0.$

Because the degree is even, at least one of $z_2, \ldots, z_{2004}$ is real, making one $b_k = 0,$ so $b_2 b_3 \cdots b_{2004} = 0.$ Thus (A) through (D) all must be $0.$

On the other hand, $\sum_{k=1}^{2004} c_k = P(1),$ and a valid polynomial such as $P(x) = x(x - 2)(x - 3) \cdots (x - 2003)\left(x + \sum_{k=2}^{2003} k\right)$ has $P(1) \ne 0.$ So only $\sum c_k$ can be nonzero.

Thus, the correct answer is E.

2004 AMC 12A Problem 23

Solution:

Problem 23 in Other Years