so one way is to factor and cross out zeroes
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
\(\frac{3a^{2}+2a-1}{a^2-1}= \frac{(3a-1)(a+1)}{(a-1)(a+1)} = (\frac{(3a-1)}{(a-1)}) (\frac{(a+1)}{(a+1)}) = (\frac{(3a-1)}{(a-1)}) (1)= \frac{(3a-1)}{(a-1)}\)
the answer is \(\frac{(3a-1)}{(a-1)}\)
