# How do you solve (x-4)(x+1)>=0?

Jun 28, 2016

$x \le - 1$ and $4 \le x$

#### Explanation:

Zeros of the function $\left(x - 4\right) \left(x + 1\right)$ are $\left\{- 1 , 4\right\}$ and they divide the real number line in three parts.

$x \le - 1$ - In this region the two binomials $\left(x + 1\right)$ and $\left(x - 4\right)$ both are negative (or zero) and hence their product is positive. So this region forms a solution.

$- 1 < x < 4$ - In this region while $\left(x + 1\right)$ is posiitive and $\left(x - 4\right)$ is negative and hence their product is negative. So this region does not form a solution.

$4 \le x$ - In this region both $\left(x + 1\right)$ and $\left(x - 4\right)$ are positive (or zero) and hence their product is positive. So this region forms a solution.

This can also be checked from the following graph.

graph{(x-4)(x+1) [-5, 5, -10, 10]}