so factor 3a^2+2a-1 write the factors (3a )(a ) now write the end factors (3a-1)(a+1) mutiply (3a^2-a+3a-a)=3a^2+2a-1 correct
factor a^2-1 difference of 2 perfect squares a^2-b^2=(a-b)(a+b) a^2-1^2=(a-1)(a+1)
so now we have [tex] \frac{3a^{2}+2a-1}{a^2-1}= \frac{(3a-1)(a+1)}{(a-1)(a+1)} = (\frac{(3a-1)}{(a-1)}) (\frac{(a+1)}{(a+1)}) = (\frac{(3a-1)}{(a-1)}) (1)= \frac{(3a-1)}{(a-1)}[/tex]
