#### Answer

$(y+2)(2y+3)(y-2)(2y+1)$

#### Work Step by Step

1. Factor the denominators.
(To factor $ax^{2}+bx+c$, find factors of $ac$ whose sum is $\mathrm{b}$.
If they exist, rewrite $bx$ and factor in pairs)
$2y^{2}+7y+6=2y^{2}+4y+3y+6$
$=2y(y+2)+3(y+2)=(y+2)(2y+3)$
$2y^{2}-3y-2=2y^{2}-4y+y-2$
$=2y(y-2)+(y-2)=(y-2)(2y+1)$
2. List the factors of the first denominator.
$ LCD=(y+2)(2y+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, we add $(y-2)$ and $(2y+1)$ to the list
$LCD=(y+2)(2y+3)(y-2)(2y+1)$
4. The product of the listed factors is the least common denominator.