# How do you solve ((x+1)(x+2))/(x-3)>0?

Feb 14, 2018

$x \in \left(- 2 , - 1\right) \cup \left(3 , \infty\right)$

#### Explanation:

Since multiplying both sides of an inequality by a positive number keeps the sign unchanged, we can rewrite
$\frac{\left(x + 1\right) \left(x + 2\right)}{x - 3} > 0$
as
${\left(x - 3\right)}^{2} \frac{\left(x + 1\right) \left(x + 2\right)}{x - 3} > {\left(x - 3\right)}^{2} \times 0 = 0$

That is

$\left(x + 1\right) \left(x + 2\right) \left(x - 3\right) > 0$

Now, the function $f \left(x\right) = \left(x + 1\right) \left(x + 2\right) \left(x - 3\right)$ has zeros at $x = - 2$, $x = - 1$ and $x = + 3$. These are the points where the function can change sign. We are, of course, looking for regions where the function is positive.

When $x < - 2$, all three factors are negative, and so the product is negative. At $x = - 2$, the function is zero - in both cases, the inequality is not satisfied.

When $- 2 < x < - 1$, $\left(x + 2\right)$ is positive, while the other two factors are negative - so our function $f \left(x\right)$ is positive!

For $- 1 < x < 3$, $\left(x + 1\right)$ and $\left(x + 2\right)$ are positive, but $\left(x - 3\right)$ is negative - leading to $f \left(x\right) < 0$

Finally, for $x > 3$, all three factors are positive - and so $f \left(x\right) > 0$

Thus, the inequality is satisfied for both $- 2 < x < - 1$ and $x > 3$

This can also be seen at a glance from the graph of $f \left(x\right)$ below :
graph{(x+1)(x-2)(x+3) [-4, 4, -2, 6]}