Question

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

Answer 1

`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

`(- + +) (+ - +) (+ + -)`

Answer 2

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]}