2001 AMC 12 Problem 23

Question

A polynomial of degree four with leading coefficient

1

and integer coefficients has two real zeros, both of which are integers. Which of the following can also be a zero of the polynomial?

\dfrac{1 + i\sqrt{11}}{2}

\dfrac{1 + i}{2}

\dfrac{1}{2} + i

1 + \dfrac{i}{2}

\dfrac{1 + i\sqrt{13}}{2}

Factor out the two integer roots: the remaining factor is

x^2 + \alpha x + \beta

with

\alpha, \beta

integers

The complex zeros are

-\dfrac{\alpha}{2} \pm \dfrac{i\sqrt{4\beta - \alpha^2}}{2};

match the real part and check that

\beta

is an integer

Writing

P(x)

= (x - r)(x - s)(x^2 + \alpha x + \beta)

with integer roots

r, s,

matching coefficients forces

\alpha

and

\beta

to be integers.

The other two zeros are

-\dfrac{\alpha}{2} \pm \dfrac{i\sqrt{4\beta - \alpha^2}}{2}.

A real part of

\dfrac{1}{2}

requires

\alpha = -1,

making the imaginary part

\dfrac{\sqrt{4\beta - 1}}{2}.

Choice A needs

\sqrt{4\beta - 1} = \sqrt{11},

i.e.

\beta = 3,

an integer, so it works. The other choices force a non-integer

\beta

(for example choice E needs

\beta = 3.5,

and choice D needs

\alpha = -2

with

\beta = \tfrac{5}{4}

).

Thus, the correct answer is A.

Problem 23 in other years