Yes, that particular polynomial expression is not readily factored other than manual "try" methods.  The factors which do evenly divide the original polynomial are "(x-2)" and "(x+2)".  This begins with a visual inspection of the polynomial and identifying the prime factors and then I like to use Algebrator's Polynomial Wizard titled "Long division" to verify that an attempted factor does evenly divide the polynomial (leaving no remainder). 

To help you clearly understand our desktop software and how appropriate it may be for your needs, please use one of the following links to obtain a free trial of the program:

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For example, choosing (x+5) as a factor does leave a remainder, yet (x-2) does evenly divide the polynomial. 

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