\(4x^3 + 16x^2 - 9x - 36 >= 0\)Find a solution for 4x^3 + 16x^2 - 9x - 36>/= 0
+128474 4x^2 + 16x^2 - 9x - 36 ≥ 0 set this to 0
4x^2 + 16x^2 - 9x - 36 = 0
Factor
4x^2(x + 4) - 9(x + 4) = 0
(x + 4) (4x^2 - 9) = 0
(x + 4)(2x + 3) (2x - 3) = 0
Set each factor to 0 and solve for x ....and x = -4, x = -3/2 and x = 3/2
Look at the graph, here : https://www.desmos.com/calculator/tiouq73sj4
Notice that the intervals that make the original inequality true are [-4, -3/2 ] and [3/2, infinity )
