x^2+3x-4
_________<0
x^2+x
This is a kinda' funky one....!!!!!
Factoring the denominator, we have x(x+1) which means that we will have vertiical asymptotes at x = -1 and x = 0
Factoring the numerator, we have (x +4)(x - 1) < 0
Setting each factor to 0, we have x = -4 and x = 1 .....so.....
Our possible answers come from these intervals (-∞, -4) (-4, -1) (-1, 0) (0, 1) (1, ∞)
Look at the graph, here.......https://www.desmos.com/calculator/fmhvy1d7ke
This rational function is >0 on these intervals (-∞, -4) (-1, 0) and (1, ∞)
And it is < 0 on (-4, -1) and (0, 1)
(x + 3)(2x - 1)(2 - x) > 0
Setting each factor to 0, we have the following possible solution intervals....
(-∞, -3), (-3, 1/2), (1/2, 2), (2, ∞ )
We can pick a point in each interval to see if that interval satisfies the inequality
Picking -5 in the first interval will make the inequality true
This lets us know that the other interval that will "work" occurs on (1/2, 2)
Then, the solutions come from (-∞, -3) and (1/2, 2)
See the graph here.....https://www.desmos.com/calculator/1l4dwzcpor