Answer
$(-4, -3/2)$ U $(3/2$, infinity)
Work Step by Step
$4x^3+16x^2-9x-36 >0$
$4x^2(x+4)-9(x+4)>0$
$(x+4)(4x^2-9)>0$
$(x+4)(2x+3)(2x-3)>0$
$(x+4)(2x+3)(2x-3)=0$
$x+4=0$
$x=-4$
$2x+3=0$
$2x=-3$
$2x/2=-3/2$
$x=-3/2$
$2x-3=0$
$2x=3$
$2x/2=3/2$
$x=3/2$
(-infinity, $-4)$
$(-4, -3/2)$
$(-3/2, 3/2)$
$(3/2$, infinity)
Let $x=-10$, $x=-2$, $x=0$, $x=5$
$x=-10$
$(x+4)(2x+3)(2x-3)>0$
$(-10+4)(2*-10+3)(2*-10-3)>0$
$-14*(-20+3)(-20-3)>0$
$-14*-17*-23 >0$
$-5474 >0$ (false)
$x=-2$
$(x+4)(2x+3)(2x-3)>0$
$(-2+4)(2*-2+3)(2*-2-3)>0$
$2*(-4+3)(-4-3)>0$
$2*-1*-7 >0$
$14 >0$ (true)
$x=0$
$(x+4)(2x+3)(2x-3)>0$
$(0+4)(2*0+3)(2*0-3)>0$
$4*3*-3>0$
$-36 >0$ (false)
$x=5$
$(x+4)(2x+3)(2x-3)>0$
$(5+4)(2*5+3)(2*5-3)>0$
$9*(10+3)(10-3)>0$
$9*13*7>0$
$819 >0$ (true)
