<s>[INST] Solve $\frac{1}{x-1}-\frac{4}{x-2}+\frac{4}{x-3}-\frac{1}{x-4}<\frac{1}{30} \forall x\in R$ I plugged this into WolframAlpha. It gave the factorized form as:
$$\frac{(x-7)(x-6)(x+1)(x+2)}{(x-4)(x-3)(x-2)(x-1)}>0$$
Once we've reached this point, I know how to find the solution for $x$. But, currently, I am having trouble converting the given question expression into the factorized form above. I started off strong with:
$$(\frac{1}{x-1}-\frac{1}{x-4})+(\frac{4}{x-3}-\frac{4}{x-2})<\frac{1}{30}$$
which simplifies to:
$$\frac{-3}{(x-1)(x-4)}+\frac{12}{(x-3)(x-2)}<\frac{1}{30}$$
which simplifies to:
$$\frac{3x^2-15x+10}{(x-1)(x-2)(x-3)(x-4)}<\frac{1}{90}$$
and now this is a dead end! As you can see, I got to the denomiator, but the numerator is not what I expected it to be. I could transfer $1/30$ to the LHS but that would bring the deadly fourth power into the numerator. If this was a competitive exam question, I would not be sitting in the hall factorizing fourth powers...

Is there a simpler method to factorize the LHS? Or is there an altogether different approach to solve this question?

 [/INST] To solve the inequality, leonbloy already gave you the best suggestion.
To reconstruct WolframAlpha answer, doesn't seem that there be a shorter way other
than the plain
$$
\eqalign{
  & {{x^{\,2}  - 5x - 2} \over {\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)\left( {x - 4} \right)}} - {1 \over {30}} =   \cr 
  &  = {{30x^{\,2}  - 150x - 60 - \left( {x^{\,4}  - 10x^{\,3}  + 35x^{\,2}  - 50x + 24} \right)} \over {30\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)\left( {x - 4} \right)}} =   \cr 
  &  =  - {{x^{\,4}  - 10x^{\,3}  + 5x^{\,2}  + 100x + 84} \over {30\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)\left( {x - 4} \right)}} =   \cr 
  &  =  - {{\left( {x + 1} \right)\left( {x + 2} \right)\left( {x - 6} \right)\left( {x - 7} \right)} \over {30\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)\left( {x - 4} \right)}} \cr} 
$$
where the factorization in the last passage is to be obtained by Ruffini's rule.
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