# How do you find the excluded values of \frac{2x+1}{x^2-x-6}?

Nov 23, 2014

Since denominators cannot be zero, we have

${x}^{2} - x - 6 \ne 0$

by factoring out,

$\left(x + 2\right) \left(x - 3\right) \ne 0 \implies x \ne - 2 , 3$

Hence, $x = - 2$ and $x = 3$ are its excluded values.

I hope that this was helpful.