# How do you simplify and find the excluded values of (9x^2 - 6x+1 )/( 12x^2 -13x+3)?

$x = \frac{3}{4}$ and $x = \frac{1}{3}$ are excluded values.
$\frac{9 {x}^{2} - 6 x + 1}{12 {x}^{2} - 13 x + 3} = \frac{\left(3 x - 1\right) \left(3 x - 1\right)}{\left(4 x - 3\right) \left(3 x - 1\right)} = \frac{3 x - 1}{4 x - 3}$
The excluded values would be the zeros of the denominator $12 {x}^{2} - 13 x + 3$, which are $x = \frac{3}{4}$ and $x = \frac{1}{3}$.