How do you solve #((x+1)(x+2))/(x-3)>0#?

1 Answer
Feb 14, 2018

Answer:

# x in (-2,-1) cup (3,infty) #

Explanation:

Since multiplying both sides of an inequality by a positive number keeps the sign unchanged, we can rewrite
#{(x+1)(x+2)}/(x-3) > 0#
as
#(x-3)^2 {(x+1)(x+2)}/(x-3) > (x-3)^2 times 0 = 0#

That is

#(x+1)(x+2)(x-3) >0 #

Now, the function #f(x) = (x+1)(x+2)(x-3) # has zeros at #x=-2#, #x=-1# and #x=+3#. These are the points where the function can change sign. We are, of course, looking for regions where the function is positive.

When #x<-2#, all three factors are negative, and so the product is negative. At #x=-2#, the function is zero - in both cases, the inequality is not satisfied.

When #-2 < x < -1 #, #(x+2)# is positive, while the other two factors are negative - so our function #f(x)# is positive!

For #-1 < x < 3#, #(x+1)# and #(x+2)# are positive, but #(x-3)# is negative - leading to #f(x)<0#

Finally, for #x > 3#, all three factors are positive - and so #f(x) > 0#

Thus, the inequality is satisfied for both #-2 <x < -1 # and #x > 3#

This can also be seen at a glance from the graph of #f(x)# below :
graph{(x+1)(x-2)(x+3) [-4, 4, -2, 6]}