Dividing by a Binomial Denominator  TOPIC_SOLVED

Quadratic equations and inequalities, variation equations, function notation, systems of equations, etc.

Dividing by a Binomial Denominator

Postby RedComet on Tue Mar 08, 2011 9:34 pm

I think I've found a mistake in the review book I'm using. Would someone care to work this problem and see if they get the same thing?

(k^3 -1) / (k - 1)

I ended up with k^2 + k + 1 as the answer, but the book gives the answer as k^2 - k - 1 - (2 / [k - 1])

Thanks for the help. :)

P.S. Awesome site. I've been using this since high school (5+ years ago) and I'm using it again in conjunction with other books to get myself ready for college level Calculus. :)
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Re: Dividing by a Binomial Denominator  TOPIC_SOLVED

Postby nona.m.nona on Tue Mar 08, 2011 11:06 pm

RedComet wrote:(k^3 -1) / (k - 1)

I ended up with k^2 + k + 1 as the answer...

This result can be obtained by factoring the difference of cubes and then canceling.

RedComet wrote:...but the book gives the answer as k^2 - k - 1 - (2 / [k - 1])

Simplify:







This does not equal the original expression, so the simplification would appear to be incorrect.
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