Answer
$\dfrac{y-1}{y+1}, y \ne-1$
Work Step by Step
By factoring both the denominator and the numerator, and then cancelling the common factors, we have
\begin{align*}
\frac{3y^2-y-2}{3y^2+5y+2}&=\frac{3y^2-3y+2y-2}{3y^2+3y+2y+2}\\
\\
&=\frac{(3y^2-3y)+(2y-2)}{(3y^2+3y)+(2y+2)}\\
\\
&= \frac{3y(y-1)+2(y-1)}{3y(y+1)+2(y+1)}\\
\\
&=\frac{(y-1)(3y+2)}{(y+1)(3y+2)}\\
\\
&=\frac{y-1}{y+1}
\end{align*}
Where we factored both polynomials by grouping.
