SOLUTION: Is my answer correct for this inequality? ((x+5)^2(8-x))/(x^2-16)≥0 Had to answer in interval notation what I got was: (-∞,-4]U[4,8) Please explain Thanks

Algebra ->  Inequalities -> SOLUTION: Is my answer correct for this inequality? ((x+5)^2(8-x))/(x^2-16)≥0 Had to answer in interval notation what I got was: (-∞,-4]U[4,8) Please explain Thanks      Log On


   

Question 948356: Is my answer correct for this inequality?
((x+5)^2(8-x))/(x^2-16)≥0
Had to answer in interval notation
what I got was: (-∞,-4]U[4,8)
Please explain
Thanks
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
%28%28x%2B5%29%5E2%288-x%29%29%2F%28x%5E2-16%29%3E=0+......multiply %288-x%29 by -1 to get -%28x-8%29, put minus in front and do not forget, sign ≥ will become £
%28-%28x%2B5%29%5E2%28x-8%29%29%2F%28x%5E2-4%5E2%29%3C=0
%28-%28x%2B5%29%5E2%28x-8%29%29%2F%28%28x-4%29%28x%2B4%29%29%3C=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:
%28x-4%29%28x%2B4%29=0
if %28x-4%29=0=>x=4
if %28x%2B4%29=0=>x=-4
also, we need to exclude values of x that make numerator equal to zero and they are
%28x%2B5%29=0=>x=-5
x-8=0 => x=8
so,since we already have excludedx=-4 and =>-4%3E-5 then x=-5 will be included in interval (-infinity, -4)
so, interval notation is:

(-infinity, -4) U (4, 8]



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