# How do you solve (x-2)(x+1)(x-5)>=0 using a sign chart?

Nov 25, 2017

Solution: $- 1 \le x < 2 \mathmr{and} x \ge 5 \mathmr{and} \left[- 1 , 2\right] \cup \left[5 , \infty\right)$

#### Explanation:

$f \left(x\right) = \left(x - 2\right) \left(x + 1\right) \left(x - 5\right) \ge 0$ . Crititical numbers

are $x = - 1 , x = 2 , x = 5$. Since at those numbers $f \left(x\right) = 0$

Sign chart:

When $x < - 1$ sign of $\left(x - 2\right) \left(x + 1\right) \left(x - 5\right)$ is  (-) * (-)* (-) = (-) ; < 0

When $- 1 < x < 2$ sign of $\left(x - 2\right) \left(x + 1\right) \left(x - 5\right)$ is  (-) * (+)* (-) = (+) ; > 0

When $2 < x < 5$ sign of $\left(x - 2\right) \left(x + 1\right) \left(x - 5\right)$ is  (+) * (+)* (-) = (-) ; < 0

When $x > 5$ sign of $\left(x - 2\right) \left(x + 1\right) \left(x - 5\right)$ is  (+) * (+)* (+) = (+) ; > 0

Solution: $- 1 \le x < 2 \mathmr{and} x \ge 5 \mathmr{and} \left[- 1 , 2\right] \cup \left[5 , \infty\right)$

graph{(x-2)(x+1)(x-5) [-40, 40, -20, 20]} [Ans]