Answer
$(2, ∞)$
Work Step by Step
$x^3+2x^2-4x-8 >0$
$x^2(x+2)-4(x+2)>0$
$(x^2-4)(x+2) >0$
$(x+2)(x-2)(x+2) >0$
$x+2 >0$
$x+2-2 > 0-2$
$x > -2$
$x-2>0$
$x-2+2>0+2$
$x>2$
We have three sections: $(-∞,-2)$, $(-2, 2)$, and $(2, ∞)$. 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 $x=-5$, $x=0$, and $x=5$
$x=-5$
$x^3+2x^2-4x-8 >0$
$(-5)^3+2(-5)^2-4(-5)-8 >0$
$-125+2*25+20-8>0$
$-125+50+12 >0$
$-63 >0$ (false)
$x=0$
$x^3+2x^2-4x-8 >0$
$0^3+2*0^2-4*0-8 >0$
$0+2*0-0-8>0$
$0+0-8>0$
$-8 >0$ (false)
$x=5$
$x^3+2x^2-4x-8 >0$
$5^3+2*5^2-4*5-8 >0$
$125+2*25-20-8>0$
$105-8+50 >0$
$147 >0 $ (true)
