How do you solve #(x-2)(x+5) >0#?

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Answer 1 · 2015-05-17T21:40:24

The signs of the two factors change at #x=-5# and #x=2#. So the sign of their product #(x-2)(x+5)# will change at those two points.

In the case #x < -5#:

#(x - 2) < 0# and #(x + 5) < 0# so #(x-2)(x+5) > 0#

In the case #x = -5#:

#(x-2)(x+5) = 0#

In the case #-5 < x < 2#:

#(x - 2) < 0# but #(x+5) > 0# so #(x-2)(x+5) < 0#

In the case #x = 2#:

#(x-2)(x+5) = 0#

In the case #x > 2#:

#(x-2) > 0# and #(x+5) > 0# so #(x-2)(x+5) > 0#

Putting these cases together #(x-2)(x+5) > 0# when #x < -5# or #x > 2#.