## Dividing by a Binomial Denominator

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

### Dividing by a Binomial Denominator

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.
RedComet

Posts: 1
Joined: Tue Mar 08, 2011 9:29 pm

### Re: Dividing by a Binomial Denominator

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:

$\left(\frac{k^2\, -\, k\, -\, 1}{1}\right)\left(\frac{k\, -\, 1}{k\, -\, 1}\right)\, -\, \frac{2}{k\, -\, 1}$

$\frac{k^3\, -\, 2k^2\, +\, 1}{k\, -\, 1}\, -\, \frac{2}{k\, -\, 1}$

$\frac{k^3\, -\, 2k^2\, -\, 1}{k\, -\, 1}$

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

Posts: 249
Joined: Sun Dec 14, 2008 11:07 pm