For what values of #x# is the relation #((-2x-10)(x-3))/((x^2+2)(x-2)^2) < 0# true?

1 Answer
Sep 20, 2017

The inequality is true for #x in (-5,+3)#
BUT your real interest is in understanding how to get this... see below.

Explanation:

Given
#color(white)("XXX")((-2x-10)(x-3))/((x^2+2)(x-2)^2) < 0#

Assuming #x!=2# (in which case the left side is undefined)
we note that both #(x^2+2)# and #(x-2)^2# must be non-negative values.

So we can simplify the given inequality by multiplying both sides by #(x^2+2)(x-2)^2#
to get
#color(white)("XXX")(-2x-10)(x-3) < 0#

There are (only) two critical values for #x# which would make the left side equal to zero,
namely #x=-5# and #x=3#

These two values divide the Domain into 3 pieces (plus the two critical points):
#{: (," | ","values"," | ",," | ","values between"," | ",," | ","values"), (x=," | ",< -5," | ", -5," | ",(-5,+3)," | ",+3," | ", >+3), ((-2x-10)(x-3)=," | ",?," | ",0," | ",?," | ",0," | ",?) :}#

We can pick any value(s) we like in the non-specific ranges and test to see if the desired inequality holds:
#{: (," | ","values"," | ",," | ","values between"," | ",," | ","values"), (x=," | ",< -5," | ", -5," | ",(-5,+3)," | ",+3," | ", >+3), ("sample value:"," | ",-6," | ",," | ",0," | ",," | ",4), ((-2x-10)(x-3)=," | ",18," | ",0," | ",-30," | ",0," | ",18), ("is this " < 0," | ","No"," | ","No"," | ","Yes"," | ","No"," | ","No") :}#

From this we see that the given inequality is only true for
#color(white)("XXX") x in (-5,+3)#