How do you solve #(x+8)(x+2)(x-3)>=0#?

1 Answer
Jun 28, 2016

Answer:

#-8 <= x <= -2# and #3<=x#

Explanation:

Zeros of the function #(x+8)(x+2)(x-3)# are #{-8,-2,3}# and they divide the real number line in four parts.

#x<-8# - In this region all binomials #(x+8)#, #(x+2)# and #(x-3)# are negative and hence their product is negative. So this region is not a solution.

#-8 <= x <= -2# - In this region while #(x+8)# is positive, #(x+2)# and #(x-3)# are negative and hence their product is positive. So this region forms a solution.

#-2 < x < 3# - In this region while #(x+8)# and #(x+2)# is positive, #(x-3)# is negative and hence their product is negative. So this region does not form a solution.

#3<=x# - In this region all binomials #(x+8)#, #(x+2)# and #(x-3)# are positive and hence their product is positive. So this region forms a solution.

This can also be checked from the following graph.

graph{(x+8)(x+2)(x-3) [-12.38, 7.62, -100, 100]}