The number of points common to the graphs of
$(x-y+2)(3x+y-4)=0 \text{    and     } (x+y-2)(2x-5y+7)=0$
is:
$\text{(A) } 2\quad \text{(B) } 4\quad \text{(C) } 6\quad \text{(D) } 16\quad \text{(E) } \infty$
By the [Zero Product Property](https://artofproblemsolving.com/wiki/index.php/Zero_Product_Property), $x-y+2=0$ or $3x+y-4=0$ in the first equation and $x+y-2=0$ or $2x-5y+7=0$ in the second equation.  Thus, from the first equation, $y = x-2$ or $y =-3x+4$, and from the second equation, $y=-x+2$ or $y = \frac{2}{5}x + \frac{7}{5}$.
If a point is common to the two graphs, then the point must be in one of the lines in the first equation as well as one of the lines in the second equation.  Since the slopes of the lines are different, none of the lines are parallel.  Thus, there are $2 \cdot 2 = \boxed{\textbf{(B) } 4}$ points of intersection in the two graphs.