Factoring A difference of fifths

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SUMMARY

The discussion centers on solving the limit of the expression (x^5-32)/(x-2) as x approaches 2, utilizing the formula for a difference of fifths. The participant successfully factored the expression and calculated the limit to be 80. Alternative methods suggested include L'Hôpital's rule and polynomial division, with the latter being recommended for those not yet familiar with derivatives. The consensus is that the difference of fifths formula is not necessary for this problem, as simpler methods exist.

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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! :)
 
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PhysicsAdvice said:
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! :)

L'hôpital's rule is probably the easiest.

Doing the polynomial division is a close second, I think.
 
having not learned derivatives yet i suppose the method i indicated is the most suitable then?
 
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}).

That seems the best method to me. You should not have to "look it up".
 

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