Question 68479This question is from textbook Glencoe Algebra 2
: We are reviewing division of polynomials in trigonometry, and for homework we have to use both long and synthetic division. One of the long division problems is (50a^2-98b^2)/(10a+14b).
I know you have to include placeholders for missing degrees, but I'm not sure how many placeholders I need because the teacher did not show us how to do long division with more than one variable.
I think it would be, showing what the placeholder is for, (50a^2+0a+0ab+98b^2+0b)/(10a+14b), but I'm not entirely sure. Is this correct and if not, how do I figure it out?
This question is from textbook Glencoe Algebra 2
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! 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.
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