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?


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.
