Posted by Aaron on .
Divide
(42b^3 + 23b^2 + 38b + 49)/ (6b + 5)

algebra ll 
David Q/R,
See my reply to your repost: unless I've slipped up myself I think you may have made a typing error with the constant term of the numerator, since 40 would make the numerator exactly divisible by (6b+5), but 49 doesn't.