Answer
$(0$, infinity)
Work Step by Step
$(x+1)^2/5x > 0$
Denominator is zero when $x=0$
$(x+1)^2/5x > 0$
$(x+1)^2/5x = 0$
$(x+1)^2*5x/5x = 0*5x$
$(x+1)^2 = 0$
$\sqrt {(x+1)^2} = \sqrt 0$
$x+1 =0$
$x=-1$
(-infinity, $-1)$
$(-1,0)$
$(0$, infinity)
Let $x=-2$, $x=-.5$, $x=1$
$x=-2$
$(x+1)^2/5x > 0$
$(-2+1)^2/5*-2 > 0$
$(-1)^2/-10 >0$
$1/-10 > 0$
$-1/10 > 0$ (false)
$x=-.5$
$(x+1)^2/5x > 0$
$(-.5+1)^2/5*-.5 > 0$
$(.5)^2/-2.5 > 0$
$.25/-2.5 > 0$
$2.5/-25 > 0$
$2.5*2/-25*2 >0$
$5/-50 >0$
$-1/10 > 0$ (false)
$x=1$
$(x+1)^2/5x > 0$
$(1+1)^2/5*1 > 0$
$2^2/5>0$
$4/5 > 0$ (true)
