Question 948356
  {{{((x+5)^2(8-x))/(x^2-16)>=0 }}}......multiply {{{(8-x)}}} by {{{-1}}} to get {{{-(x-8)}}}, put minus in front and do not forget, sign ≥ will become £

  {{{(-(x+5)^2(x-8))/(x^2-4^2)<=0}}}

 {{{(-(x+5)^2(x-8))/((x-4)(x+4))<=0}}}

as you know, denominator cannot be equal to zero; so, we need to exclude values of {{{x}}} that make denominator equal to zero and they are:

{{{(x-4)(x+4)=0}}}
if {{{(x-4)=0}}}=>{{{x=4}}}
if {{{(x+4)=0}}}=>{{{x=-4}}}

also, we need to exclude values of {{{x}}} that make numerator equal to zero and they are
{{{(x+5)=0}}}=>{{{x=-5}}}
{{{x-8=0}}} => {{{x=8}}}

so,since we already have  excluded{{{x=-4}}} and =>{{{-4>-5}}} then {{{x=-5}}} will be included in interval ({{{-infinity}}}, {{{-4}}})

so,  interval notation is:
 
({{{-infinity}}}, {{{-4}}}) U ({{{4}}}, {{{8}}}]



{{{drawing( 600, 600, -20, 20, -25, 25,  
blue(line(-4,25,-4,-25)),blue(line(4,25,4,-25)),blue(line(8,25,8,-25)),
graph( 600, 600, -20, 20, -25, 25,0, ((x+5)^2(8-x))/(x^2-16))) }}}

so, your solution is part of graph above x-axis, left from blue line passing through {{{-4}}} and between blue lines passing through {{{4}}} and {{{8}}}