Factor: x^(2/5) - 3x^(1/5) - 4 = 0
Use a similar idea to that suggested here: http://www.mathhelpforum.com/math-he...lynomials.html
You have to realise that it's a quadratic.
Technically all methods for factorising involve this "substitution" - it's just whether you want to make it look a bit more pleasing to the eyes.
You could just as easily write it as
$\displaystyle (x^{\frac{1}{5}})^2 - 3x^{\frac{1}{5}} -4$
which becomes $\displaystyle (x^{\frac{1}{5}} - 4)(x^{\frac{1}{5}} + 1)$.