2001 AMC 12 Problem 23

Problem 23 of 25HarderAlgebra

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?

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Solution

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.