jmc

If $x^2+tx-10=(x+a)(x+b)$, then $$x^2+tx-10=x^2+ax+bx+ab=x^2+(a+b)x+ab.$$Therefore, we must have $ab=-10$, and for any such $a$ and $b$, we have $t=a+b$. Our possibilities are as follows: $$\begin{array}{ccc}a& b& a+b\\ -1& 10& 9\\ -2& 5& 3\\ -5& 2& -3\\ -10& 1& -9\end{array}$$The product of these possible values of $t=a+b$ is $(9)(3)(-3)(-9)=27^2=\boxed{729}$.