$29 \cdot 50$ If $x^{2}+3 x y+x+m y-m$ has two linear factors with integer coefficients in $x, y$, then the set of values of $m$ is
(A) $\{0,12,-12\}$.
(B) $\{0,12\}$.
(C) $\{12,-12\}$.
(D) $\{12\}$.
(E) $\{0\}$.
(15th American High School Mathematics Examination, 1964)
［Solution］Assume the original quadratic expression has factors $(x+a y+b)$ and $(x+c y+d)$. Expanding their product and comparing coefficients with the given expression, we have
$$
\left\{\begin{array}{l}
a+c=3, \\
b+d=1, \\
a d+b c=m, \\
b d=-m, \\
a c=0 .
\end{array}\right.
$$

From (1) and (3), we get $c=0, a=3$ or $a=0, c=3$.
Substituting $c=0, a=3$ into (3) yields $3 d=m$,
Substituting (6) into (4) yields $3 d=-b d$.
If $d=0$, then $m=0$;
If $d \neq 0$, from (7) we have $b=-3$.
Then from (2) we have $d=4$, substituting into (4) yields $m=12$.
By symmetry, substituting $a=0, c=3$ yields the same result.
In summary,
$m=0$ or 12.

Therefore, the answer is $(B)$.