Question 68479
One of the long division problems is 
(50a^2-98b^2)/(10a+14b)
=[2(25a^2-49b^2)] / [2(5a+7b)]
The 2's cancel and the numberator factors as follows:
=[(5a-7b)(5a+7b)]/(5a+7b)
The (5a+7b) factors cancel.
=5a-7b
Cheers,
Stan H.