2004 AMC 12A Problem 23

Question

A polynomial

\begin{aligned} &P(x) = c_{2004} x^{2004} + c_{2003} x^{2003} \\ &\quad {}+ \cdots + c_1 x + c_0 \end{aligned}

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

Since

z_1 = 0

is a root,

c_0 = P(0) = 0;

nonreal roots come in conjugate pairs

\sum c_k = P(1),

which is not forced to be zero; check each other option must vanish

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

\cdot (x - 2003)

\cdot \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.

Problem 23 in other years