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?

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

$\dfrac{1 + i}{2}$

$\dfrac{1}{2} + i$

$1 + \dfrac{i}{2}$

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

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.

2001 AMC 12 Problem 23

Solution:

Problem 23 in Other Years