Answer
$\dfrac{y-1}{y+1}, \space y\ne\dfrac{-2}{3}$.
Work Step by Step
Factor $3y^2-y-2$:-
With $a=3$ and $c=-2$, then $ac=3(-2)=-6$.
The factors of $-6$ whose sm is $-1$ (the value of $b$) are $-3$ and $2$.
Thus, rewrite $-y$ as $-3y+2y$ to obtain:
$3y^2-y-2=3y^2-3y+2y-2$
Group the terms.
$=(3y^2-3y)+(2y-2)$
Factor out the GCF in each group.
$=3y(y-1)+2(y-1)$
Factor out $(y-1)$.
$=(y-1)(3y+2)$
Factor $3y^2+5y+2$:-
With $a=3$ and $c=2$, then $ac=3(2)=6$.
The factors of $6$ whose sum is $5$ are $3$ and $2$.
Thus, rewrite $5y$ as $3y+2y$.
$3y^2+5y+2=3y^2+3y+2y+2$
Group the terms.
$=(3y^2+3y)+(2y+2)$
Factor out the GCF in each group.
$=3y(y+1)+2(y+1)$
Factor out $(y+1)$.
$=(y+1)(3y+2)$
So now we have .
$\dfrac{3y^2-y-2}{3y^2+5y+2}=\dfrac{(y-1)(3y+2)}{(y+1)(3y+2)}$
Cancel the common factor $3y+2$ to obtain:.
$=\dfrac{y-1}{y+1}, y\ne-\frac{2}{3}$
Hence, the lowest term is $\dfrac{y-1}{y+1}, y\ne-\frac{2}{3}.$.
