Question 1006083
{{{x(16-x^2)>=0}}}
{{{x(4+x)(4-x)>=0}}}


Critical x values:  -4,0,4;


?  (-infinity,-4] ?
(-)(-)(+)>=0
TRUE


[-4,0] ?
Try -1.
(-1)(4-1)(4-(-1))=(-1)(3)(5)>=0
FALSE


[0,4] ?
Try 1
1*(4+1)(4-1)=1*5*3>=0
TRUE


[4, infinity) ?
5*(4+5)(4-5)=Negative>=0
FALSE


Solution set is  (-infinity, -4] U [0,4].
Choice E.