Help solving the following inequality $$
\frac{x|x + 1|(x + 2)}{|x - 1|} \ge 0
$$
My idea was to multiply the denominator to $0$ and $| x + 1 |$ to solve it  in two cases  (when is positive and when is negative). No luck
 A: Hint:
$$\frac{x|x + 1|(x + 2)}{|x - 1|} =\underbrace{\frac{|x + 1|}{|x - 1|}}_+.\left(x^2+2x\right)$$
A: There are nine cases to check out, to cover all possible significant values of x; four are easy,
The LHS is zero  for $x=-2, -1, \text{and } 0$;  the inequality is true
The LHS is undefined for $x=1$;  the inequality is false
Now for the remaining five cases:in all of them the absolute value terms are positive, and can be ignored as to their effect on the sign of the LHS.  Only the $x$ and $x+2$ terms are relevant.
$x<-2$:  Both relevant terms are negative;  the product is positive and the inequality is true.
$-2 < x<-1$:  The two relevant terms differ in sign; the product is negative and the inequality is false. 
$-1 < x<0$:  The two relevant terms differ in sign; the product is negative and the inequality is false. 
$0 < x<1$:  The two relevant terms are positive; the product is positive and the inequality is true. 
$x>1$:  The two relevant terms are positive; the product is positive and the inequality is true
