How do you solve #(x+4)(x-2)(x-7)>0#?

2 Answers
Jul 21, 2016

Answer:

#x in (-4, 7) and x > 7#.

Explanation:

The answer was obtained by considering the scheme of signs of the

factors for which the product > 0. The favorable schemes are

#(+ - -) (- + -) (- - +)# and

# (- + +) (+ - +) (+ + -)#

Jul 21, 2016

Answer:

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

Explanation:

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]}