jmc

If $x^2+kx+15=(x+a)(x+b)$, then $$x^2+kx+15=x^2+ax+bx+ab=x^2+(a+b)x+ab.$$Therefore, we must have $ab=15$, and for any such $a$ and $b$, we have $k=a+b$. There are four pairs of integers that multiply to 15. They are 1 and 15 (which give $k=16$), 3 and 5 (which give $k=8$), $-1$ and $-15$ (which gives $k=-16$), and -3 and -5 (which give $k=-8$). The product of these four possible values of $k$ is $$\begin{array}{rl}(16)(8)(-16)(-8)& =(2^4)(2^3)(-2^4)(-2^3)\\ & =2^{4+3+4+3}\\ & =2^{14}\\ & =2^{10}\cdot 2^4=(1024)(16)=\boxed{16384}.\end{array}$$