(2x)/(2x^2+5x+2) > 1/(x+1) where x!=-1,-1/2,-2
(2x)/(2x^2+5x+2) -1/(x+1) >0
(2x(x+1)-(2x^2+5x+2))/((x+1)(2x^2+5x+2)) >0
(2x^2+2x-2x^2-5x-2)/((x+1)(2x^2+5x+2)) >0
(-3x-2)/((x+1)(2x^2+5x+2)) >0
((x+1)(2x^2+5x+2))^2times(-3x-2)/((x+1)(2x^2+5x+2)) >0times((x+1)(2x^2+5x+2))^2
color (red) ("you need to multiply both sides by the denominator squared because you don't know whether your denominator is a positive or negative number. If it was negative, then your inequality sign would change."
(-3x-2)(x+1)(2x^2+5x+2) >0
(-3x-2)(x+1)(2x+1)(x+2)>0
-(3x+2)(x+1)(2x+1)(x+2)>0
graph{(-3x-2)(x+1)(2x+1)(x+2) [-10, 10, -5, 5]}
looking at the graph, we can see that
-2/3 < x < -1/2 and -2 < x <-1 only
