jmc

If $r$ and $s$ are the integer zeros, the polynomial can be written in the form $$P(x)=(x-r)(x-s)(x^2+\alpha x+\beta ).$$The coefficient of $x^3$, $\alpha -(r+s)$, is an integer, so $\alpha$ is an integer. The coefficient of $x^2$, $\beta -\alpha (r+s)+rs$, is an integer, so $\beta$ is also an integer. Applying the quadratic formula gives the remaining zeros as $$\frac{1}{2}(-\alpha \pm \sqrt{{\alpha }^2-4\beta })=-\frac{\alpha }{2}\pm i\frac{\sqrt{4\beta -{\alpha }^2}}{2}.$$Answer choices (A), (B), (C), and (E) require that $\alpha =-1$, which implies that the imaginary parts of the remaining zeros have the form $\pm \sqrt{4\beta -1}/2$. This is true only for choice $\boxed{\text{(A)}}$. Note that choice (D) is not possible since this choice requires $\alpha =-2$, which produces an imaginary part of the form $\sqrt{\beta -1}$, which cannot be $\frac{1}{2}$.