Answer
$(y-3)(y+3)^{2}(2y-1)$
Work Step by Step
1. Factor the denominators.
1st denominator:
$y^{2}-9=y^{2}-3^{2}=(y-3)(y+3)$
2nd denominator
$ y^{2}+6y+9=(y)^{2}+2(y)(3)+(3)^{2}$
$=$square of a sum $=(y+3)^{2}=(y+3)(y+3)$
3rd denominator
(To factor $ax^{2}+bx+c$, find factors of $ac$ whose sum is $\mathrm{b}$.
If they exist, rewrite $bx$ and factor in pairs)
$2y^{2}+5y-3=2y^{2}+6y-y-3$
$=2y(y-3)-(y-3)=(y-3)(2y-1)$
2. List the factors of the first denominator.
$ LCD=(y-3)(y+3)...\quad$ (for now,)
3. For each next denominator, add to the list any factors that do not yet appear in the list.
From the 2nd denominator,
one $(y+3)$ is in the list, we add the other $(y+3)$ to the list
$ LCD=(y-3)(y+3)(y+3)...\quad$ (for now,)
From the 3rd denominator,
$(y-3)$ is in the list, we add $(2y-1)$ to the list
$LCD=(y-3)(y+3)(y+3)(2y-1)$
4. The product of the listed factors is the least common denominator.
$LCD=(y-3)(y+3)^{2}(2y-1)$
