How do you solve #(x + 3) ( x + 6) \leq 0#?

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Answer 1 · 2017-11-28T21:59:31

#-6<= x <= -3#

#(x+3)(x+6)<=0#

Let's start by distributing the left side

#(x)(x)+(x)(6)+(3)(x)+(3)(6)<=0#

#x^2 + 6x + 3x + 18 <=0#

#x^2 + 9x + 18 <=0#

Now we need to find the critical points of the inequality

#x^2 + 9x + 18 = 0#

Now we need to factorize the left side

#(x+3)(x+6)=0#

Set factors equal to #0#

#x+3 =0 or x+6=0#

#x=0-3 or x= 0-6#

#x = -3 or x=-6#

Now we need to check the intervals in between critical points.

We have:
#x<=-6# It doesn't work in the original inequality

#-6<=x<=-3# This works!

#x>=-3# This doesn't work.

Thus, the correct answer is

#-6 <=x<=-3#