How do you solve #(x-4)(x+1)>=0#?

1 Answer
Jun 28, 2016

Answer:

# x <= -1# and #4<=x#

Explanation:

Zeros of the function #(x-4)(x+1)# are #{-1,4}# and they divide the real number line in three parts.

#x<=-1# - In this region the two binomials #(x+1)# and #(x-4)# both are negative (or zero) and hence their product is positive. So this region forms a solution.

#-1 < x < 4# - In this region while #(x+1)# is posiitive and #(x-4)# is negative and hence their product is negative. So this region does not form a solution.

#4 <= x # - In this region both #(x+1)# and #(x-4)# are positive (or zero) and hence their product is positive. So this region forms a solution.

This can also be checked from the following graph.

graph{(x-4)(x+1) [-5, 5, -10, 10]}