## Intermediate Algebra for College Students (7th Edition)

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