How do you solve and write the following in interval notation: #-(x + 6)(x + 2)(x - 2) >0#?

1 Answer
Oct 17, 2017

Answer:

#(x < -6)# or #(-2 < x < 2)#

Explanation:

The expression #color(blue)(-(x+6)(x+2)(x-2))#
is obviously equal to #color(blue)(0)# at #x in {color(magenta)(-6),color(magenta)(-2),color(magenta)(+2)}#

This gives us the critical points where the value of the expression might change between positive and negative (where the inequality #< 0# might apply).

We can select any other values below this set, between values in this set, and above this set to determine if the expression is #> 0# between these critical values.
For example:
#{: ("critical value at "x=,,color(magenta)(-6),,color(magenta)(-2),,color(magenta)(+2),), ("sample value for "x,=color(green)(-7),,color(green)(-4),,color(green)0,,color(green)(+4)) :}#

Evaluating at these sample values:
#{: ("when "x=," | ",-(x+6)(x+2)(x-2)=," | ","is this "> 0?), (color(green)(-7)," | ",+45," | ",color(red)("Yes")), (color(green)(-4)," | ",-24," | ","No"), (color(white)(.)color(green)0," | ",+24," | ",color(red)("Yes")), (color(green)(+4)," | ",-45," | ","No") :}#

This gives us that
for #color(red)(x < -6)#
#color(white)("XXX")color(red)(-(x+6)(x+2)(x-2) > 0)#
for #-6 < x < -2#
#color(white)("XXX")-(x+6)(x+2)(x-2) cancel(>) 0#
for #color(red)(-2 < x < +2)#
#color(white)("XXX")color(red)(-(x+6)(x+2)(x-2) > 0)#
for #x > +2#
#color(white)("XXX")-(x+6)(x+2)(x-2) cancel(>) 0#

[note that at #x=-6 or -2 or +2# the expression is #=0# and
therefore it can not be #> 0# for these values.