# Factoring A difference of fifths

1. Sep 12, 2011

One of the homework questions i did tonight was (x^5-32)/(x-2), where i had to find the limit when x approaches two. In order to do this i looked up the formula to use for a difference of fifths and was able to solve the question by factoring out the denominator and putting two in for x, giving me an answer of 80.

I am wondering if there is a much simpler way of solving this which i foolishly missed as we are definitely not expected to know the formula for a difference of fifths and i'm not sure it is easily derived. If anyone can spot one or argue that there isn't one that would be great! :)

2. Sep 12, 2011

### Hurkyl

Staff Emeritus
L'hôpital's rule is probably the easiest.

Doing the polynomial division is a close second, I think.

3. Sep 12, 2011

It is generally true that $x^n- y^n= (x- y)(x^{n-1}+ x^{n-2}y+ \cdot\cdot\cdot+ xy^{n-2}+ y^{n-1})$.