We have #(x+4)(x-2)(x-7)>0#

The three zeros of function #(x+4)(x-2)(x-7)# divide real number line in four parts

**(1)** #x<-4# - Here all the terms are negative, hence #f(x)# is positive. Hence, this does not form part of solution.

**(2)** #-4 < x < 2# - Here while first term #(x+1)# is positive, other two terms are negative and hence #f(x)# is positive. Hence, this forms part of solution.

**(3)** #2 < x < 7# - Here while first two terms #(x+4)# and #(x-2)# are positive, while #(x-7)# is negative and hence #f(x)# is negative. Hence, this does not form part of solution.

**(4)** #7 < x # - Here all the three term are positive and hence #f(x)# is positive. Hence, this forms part of solution.

Hence solution set for #(x+4)(x-2)(x-7)>0# is #-4 < x < 2# and #7 < x#.

graph{(x+4)(x-2)(x-7) [-10, 10, -160, 160]}