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

Dec 20, 2016

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

#### Explanation:

$\frac{{x}^{2} - 3 x + 2}{{x}^{2} - 2 x - 3} = \frac{\left(x - 1\right) \left(x - 2\right)}{\left(x - 3\right) \left(x + 1\right)}$

The individual linear factors change sign when $x \in \left\{- 1 , 1 , 2 , 3\right\}$

So the only points where the quotient can change sign are those.

In addition, note that each linear factor only occurs once. So the quotient does change sign at each of those points.

When $x > 3$, all of the linear factors are positive and so the quotient is positive. Hence its sign on each of the intervals between the points we have identified alternates as follows:

$\left(3 , \infty\right) : \text{ } +$

$\left(2 , 3\right) : \text{ } -$

$\left(1 , 2\right) : \text{ } +$

$\left(- 1 , 1\right) : \text{ } -$

$\left(- \infty , - 1\right) : \text{ } +$

So the solution set is:

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

graph{(y-(x^2-3x+2)/(x^2-2x-3)) = 0 [-8.915, 11.085, -3.98, 6.02]}