For how many values of the constant $k$ will the polynomial $x^{2}+kx+36$ have two distinct integer roots?
$\textbf{(A)}\ 6 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 14 \qquad\textbf{(E)}\ 16$
Let $p$ and $q$ be the roots of $x^{2}+kx+36.$ By [Vieta's Formulas](https://artofproblemsolving.com/wiki/index.php/Vieta%27s_Formulas), we have $p+q=-k$ and $pq=36.$
It follows that $p$ and $q$ must be distinct factors of $36.$ The possibilities of $\{p,q\}$ are \[\pm\{1,36\},\pm\{2,18\},\pm\{3,12\},\pm\{4,9\}.\]
Each unordered pair gives a unique value of $k.$ Therefore, there are $\boxed{\textbf{(B) }8}$ values of $k,$ namely $\pm37,\pm20,\pm15,\pm13.$
~stevens0209 ~MRENTHUSIASM ~$\color{magenta} zoomanTV$