## Algebra 1: Common Core (15th Edition)

Simplified expression: $\frac{2(a-1)}{3(a+1)}$ Excluded values: $-1$ and $1$.
$\frac{2a^2-4a+2}{3a^2-3}$ In a rational expression, excluded values are any real numbers that make the denominator equal to zero. In this case, any solutions to the equation $3a^2-3=0$ would be excluded values. $3a^2-3=0$ $3(a^2-1)=0$ $3(a-1)(a+1)=0$ $a-1=0$ or $a+1=0$ $a=1$ or $a=-1$ Therefore, -1 and 1 are two excluded values. Now, we can find the simplified expression: $\frac{2a^2-4a+2}{3a^2-3}$ $=\frac{2(a^2-2a+1)}{3(a-1)(a+1)}$ $=\frac{2(a-1)^2}{3(a-1)(a+1)}$ $=\frac{2(a-1)}{3(a+1)}$