# How do you solve (x^2-x-2)/(x^2-4x+3)>0?

Jul 11, 2016

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

#### Explanation:

$\frac{{x}^{2} - x - 2}{{x}^{2} - 4 x + 3} = \frac{\left(x - 2\right) \left(x + 1\right)}{\left(x - 1\right) \left(x - 3\right)}$

So the numerator changes sign at $x = - 1$ and $x = 2$ while the denominator changes sign at $x = 1$ and $x = 3$

Each change of sign (numerator or denominator) changes the sign
of the quotient.

So the sign reverses each time we move to the next interval in the sequence:

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

For large positive values of $x$, all of the linear factors are positive, so $\frac{{x}^{2} - x - 2}{{x}^{2} - 4 x + 3} > 0$.

So the signs of the quotient in the $5$ intervals listed are:

$+ - + - +$

So the quotient is positive in $\left(- \infty , - 1\right) \cup \left(1 , 2\right) \cup \left(3 , \infty\right)$

graph{(x^2-x-2)/(x^2-4x+3) [-10, 10, -5, 5]}