The expression $21x^2 +ax +21$ is to be factored into two linear prime binomial factors with integer coefficients. This can be done if $a$ is:
$\textbf{(A)}\ \text{any odd number} \qquad\textbf{(B)}\ \text{some odd number} \qquad\textbf{(C)}\ \text{any even number}$
$\textbf{(D)}\ \text{some even number} \qquad\textbf{(E)}\ \text{zero}$
We can factor $21x^2 + ax + 21$ as $(7x+3)(3x+7)$, which expands to $21x^2+42x+21$. So the answer is $\textbf{(D)}\ \text{some even number}$