Posted by **Aaron** on Friday, May 4, 2012 at 9:26pm.

Divide

(42b^3 + 23b^2 + 38b + 49)/ (6b + 5)

- algebra ll -
**David Q/R**, Saturday, May 5, 2012 at 3:04pm
See my reply to your re-post: 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.

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