# How do you solve ((x+3)(x-4))/(x+5)>=0?

Oct 14, 2016

$x \in \left(- 5 , - 3\right] \cup \left[4 , \infty\right)$

#### Explanation:

$f \left(x\right) = \frac{\left(x + 3\right) \left(x - 4\right)}{x + 5}$

Note that $f \left(x\right)$ is the product of $3$ terms:

$\left(x + 3\right)$, $\left(x - 4\right)$ and $\frac{1}{x + 5}$

These terms change sign when $x = - 5$, $x = - 3$ and $x = 4$. Hence $f \left(x\right)$ changes sign at each of these values of $x$.

When $x = - 5$ the denominator is zero, so $f \left(x\right)$ is undefined. So $x = - 5$ is not part of the solution space.

When $x = - 3$ or $x = 4$ the numerator is $0$ and the inequality is satisfied. So the solution space includes $\left\{- 3 , 4\right\}$

The three values of $x$ cut the Real line into $4$ intervals in which the sign of $f \left(x\right)$ does not change, namely:

$\left(- \infty , - 5\right)$, $\left(- 5 , - 3\right)$, $\left(- 3 , 4\right)$, $\left(4 , \infty\right)$

For large positive values of $x$ we can see that $f \left(x\right) > 0$. So the signs of the left hand side in each of these intervals is:

$\left(- \infty , - 5\right) \text{: } f \left(x\right) < 0$

$\left(- 5 , - 3\right) \text{: } f \left(x\right) > 0$

$\left(- 3 , 4\right) \text{: } f \left(x\right) < 0$

$\left(4 , \infty\right) \text{: } f \left(x\right) > 0$

Remembering that $\left\{- 3 , 4\right\}$ is part of the solution space, the complete solution space is:

$\left(- 5 , - 3\right] \cup \left[4 , \infty\right)$

graph{((x+3)(x-4))/(x+5) [-11.83, 8.17, -30.8, 29.2]}