Question 1127489
<pre>
She only gave one solution.  Also she made some illegal assumptions. There
are 8 altogether, although there are really only 4, because the two parentheses
on the right can be in reverse order.  That would be a, b and c trading places
with d, e and f, respectively. For instance, solutions 1 and 3 below are
essentially the same.  Here are all 8 possibilities (4 essentially): 

1.   x² - y² + x + 5y - 6 = (-x - y + 2)(-x + y - 3), in which
     a=-1, b=-1, c=2, d=-1, e=1, f=-3, m=1

2.   x² - y² - x + 5y - 6 = (-x - y + 3)(-x + y - 2), in which
     a=-1, b=-1, c=3, d=-1, e=1, f=-2, m=-1

3.   x² - y² + x + 5y - 6 = (-x + y - 3)(-x - y + 2), in which
     a=-1, b=1, c=-3, d=-1, e=-1, f=2, m=1

4.   x² - y² - x + 5y - 6 = (-x + y - 2)(-x - y + 3), in which
     a=-1, b=1, c=-2, d=-1, e=-1, f=3, m=-1

5.   x² - y² - x + 5y - 6 = (x - y + 2)(x + y - 3), in which
     a=1, b=-1, c=2, d=1, e=1, f=-3, m=-1

6.   x² - y² + x + 5y - 6 = (x - y + 3)(x + y - 2), in which
     a=1, b=-1, c=3, d=1, e=1, f=-2, m=1

7.   x² - y² - x + 5y - 6 = (x + y - 3)(x - y + 2), in which
     a=1, b=1, c=-3, d=1, e=-1, f=2, m=-1

8.   x² - y² + x + 5y - 6 = (x + y - 2)(x - y + 3), in which
     a=1, b=1, c=-2, d=1, e=-1, f=3, m=1

Edwin</pre>