Finding number of real roots The question is 


  Find the number of real roots of the equation
  $(x-1)x(x+1)(x+2)-1=0$


I simply tried to simplify the equation but it transforms into biquadratic which becomes really tough to conclude anything from it.My second approach was :-
$y=(x^2-1)  ;$
$y=\frac{1}{(x^2+3x+2)}$
Then i tried to find their intersection but i failed 
Please help to solve the question without using any tools (like wolfram) other than pure mathematics since it is subjective mathematical paper's question which needs a proper solution.
The function is symmetric about $x = -{1 \over 2}$ since the zeroes of $(x+2)(x+1)x(x-1)$ are at $-2,-1,0,$ and $1$. So letting $y = x + {1 \over 2}$ the polynomial is 
$$(y-{1 \over 2})(y + {1 \over 2})(y - {3 \over 2})(y + {3 \over 2}) - 1$$
$$ = y^4 - {5 \over 2}y^2 -  {7 \over 16}$$ 
This is a quadratic equation in $y^2$, so you can find all possible $y$ easily, and from that the number of solutions $x$.
Factorize 

$(x^2+x)(x^2+x-2)-1=0$

$(x^2+x)^2-2(x^2+x)-1=0$ 

$(x^2+x-1)^2=2$