Question 121749
If {{{x + 3< 0}}} or {{{x + 8< 0}}}, then {{{(x + 3)(x + 8) < 0}}}
So, the solutions are:

{{{x + 3< 0}}}…=>

{{{x < -3}}}…=>…. Or  "{x | x is a real number, x < –3}", 

"Interval notation" writes the solution as an interval  "{{{-infinity}}}, {{{-3}}} ",

and

{{{x + 8< 0}}}…=>


{{{x < -8}}}…=>Or  "{x | x is a real number, x < –8}",

"Interval notation" writes the solution as an interval  " {{{-infinity}}}, {{{-8}}} "


here is the graph:

{{{ (x + 3)(x + 8) < 0 }}}

{{{ x^2 + 8x  + 3x + 24< 0 }}}

{{{ x^2 + 11x + 24< 0 }}}

Graph it as an equation, then shade part where {{{ x^2 + 8x  + 3x + 24< 0 }}}


*[invoke Plot_a_graph_of_functions "x^2 + 11x + 24 ", -15, 10, -20, 20]