2001 AMC 12 Problems/Problem 23

Problem

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?

$\text{(A) }\frac {1 + i \sqrt {11}}{2} \qquad \text{(B) }\frac {1 + i}{2} \qquad \text{(C) }\frac {1}{2} + i \qquad \text{(D) }1 + \frac {i}{2} \qquad \text{(E) }\frac {1 + i \sqrt {13}}{2}$

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 $\frac {1 \pm i \sqrt {11}}{2}$ .)

2001 AMC 12 (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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2001 AMC 12 Problems/Problem 23

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