Let's find the quotient of the given rational expressions step by step:
Given:
[tex]\[
\frac{2y^2 - 6y - 20}{4y + 12} \div \frac{y^2 + 5y + 6}{3y^2 + 18y + 27}
\][/tex]
First, we recall that dividing by a fraction is equivalent to multiplying by its reciprocal:
[tex]\[
\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}
\][/tex]
Applying this to our expressions:
[tex]\[
\frac{2y^2 - 6y - 20}{4y + 12} \times \frac{3y^2 + 18y + 27}{y^2 + 5y + 6}
\][/tex]
Next, we factor the polynomials where possible:
1. For the numerator [tex]\(2y^2 - 6y - 20\)[/tex]:
[tex]\[
2y^2 - 6y - 20 = 2(y^2 - 3y - 10) = 2(y - 5)(y + 2)
\][/tex]
2. For the denominator [tex]\(4y + 12\)[/tex]:
[tex]\[
4y + 12 = 4(y + 3)
\][/tex]
3. For the numerator [tex]\(3y^2 + 18y + 27\)[/tex]:
[tex]\[
3y^2 + 18y + 27 = 3(y^2 + 6y + 9) = 3(y + 3)^2
\][/tex]
4. For the denominator [tex]\(y^2 + 5y + 6\)[/tex]:
[tex]\[
y^2 + 5y + 6 = (y + 2)(y + 3)
\][/tex]
Substitute the factored forms back into our expression:
[tex]\[
\frac{2(y - 5)(y + 2)}{4(y + 3)} \times \frac{3(y + 3)^2}{(y + 2)(y + 3)}
\][/tex]
Cancel out the common factors:
1. [tex]\((y + 2)\)[/tex] in the numerator and denominator.
2. One [tex]\((y + 3)\)[/tex] in the numerator and denominator.
This leaves us with:
[tex]\[
\frac{2(y - 5)}{4} \times \frac{3(y + 3)}{(y + 2)}
\][/tex]
Simplify the multiplication:
[tex]\[
\frac{2(y - 5) \cdot 3(y + 3)}{4 \cdot (y + 3)} = \frac{2 \cdot 3(y - 5)(y + 3)}{4(y + 3)}
\][/tex]
Further, we can cancel another [tex]\((y + 3)\)[/tex] term:
[tex]\[
\frac{6(y - 5)}{4} = \frac{3(y - 5)}{2}
\][/tex]
Thus, the quotient is:
[tex]\[
\boxed{\frac{3(y - 5)}{2}}
\][/tex]
Therefore, the correct answer is:
[tex]\[
\boxed{\frac{3(y-5)}{2}}
\][/tex]
