Answer
$(-4, -3/2)$ U $(3/2, ∞)$
Work Step by Step
$4x^3+16x^2-9x-36 >0$
$4x^2(x+4)-9(x+4) > 0$
$(4x^2-9)(x+4)>0$
$(2x+3)(2x-3)(x+4) > 0$
$2x+3 >0$
$2x+3-3 >0-3$
$2x > -3$
$2x/2 > -3/2$
$x > -3/2$
$2x-3 >0$
$2x-3+3 >0+3$
$2x > 3$
$2x/2 > 3/2$
$x > 3/2$
$x+4 >0$
$x+4-4 > 0-4$
$x > -4$
We have four sections: $(-∞,-4)$, $(-4, -3/2)$, $(-3/2, 3/2)$, and $(3/2, ∞)$. We need to test one value for $x$ in each section to determine if the section would be a solution set. Since we have the $>$ sign, we exclude the end points and use parentheses instead of brackets.
Let $x=-5$, $x=-2$, $x=0$, and $x=2$
$x=-5$
$4(-5)^3+16(-5)^2-9(-5)-36 >0$
$4*-125+16*25+45-36>0$
$-500+400+9 >0$
$-91 > 0$ (false)
$x=-2$
$4x^3+16x^2-9x-36 >0$
$4(-2)^3+16(-2)^2-9(-2)-36 >0$
$4*-8 +16*4+18-36 >0$
$-32+64-18 > 0$
$14 >0$ (true)
$x=0$
$4x^3+16x^2-9x-36 >0$
$4*0^3+16*0^2-9*0-36 >0$
$4*0+16*0-0-36>0$
$0+0-36 >0$
$-36 >0$ (false)
$x=2$
$4x^3+16x^2-9x-36 >0$
$4*2^3+16*2^2-9*2-36 >0$
$4*8+16*4-18-36 >0$
$32+64-54 >0$
$42 >0$ (true)
