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

Jul 21, 2016

$x \in \left(- 4 , 7\right) \mathmr{and} x > 7$.

#### Explanation:

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

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

$\left(+ - -\right) \left(- + -\right) \left(- - +\right)$ and

$\left(- + +\right) \left(+ - +\right) \left(+ + -\right)$

Jul 21, 2016

Solution set for $\left(x + 4\right) \left(x - 2\right) \left(x - 7\right) > 0$ is $- 4 < x < 2$ and $7 < x$.

#### Explanation:

We have $\left(x + 4\right) \left(x - 2\right) \left(x - 7\right) > 0$

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

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

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

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

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

Hence solution set for $\left(x + 4\right) \left(x - 2\right) \left(x - 7\right) > 0$ is $- 4 < x < 2$ and $7 < x$.

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