Alright, let's solve the quotient step-by-step.
The given expression is:
[tex]\[
\frac{2 y^2-6 y-20}{4 y+12} \div \frac{y^2+5 y+6}{3 y^2+18 y+27}
\][/tex]
### Step 1: Rewrite the division as a multiplication of the reciprocal
The division of two fractions can be converted to multiplication by taking the reciprocal of the second fraction. So, we rewrite our expression as:
[tex]\[
\frac{2 y^2-6 y-20}{4 y+12} \times \frac{3 y^2+18 y+27}{y^2+5 y+6}
\][/tex]
### Step 2: Factorize Numerator and Denominator Expressions
To simplify further, let's factorize the expressions where possible.
1. Factorizing \(2 y^2 - 6 y - 20\):
[tex]\[
2 y^2 - 6 y - 20 = 2 (y^2 - 3y - 10) = 2 (y-5)(y+2)
\][/tex]
2. Factorizing \(4 y + 12\):
[tex]\[
4 y + 12 = 4(y + 3)
\][/tex]
3. Factorizing \(y^2 + 5 y + 6\):
[tex]\[
y^2 + 5 y + 6 = (y+2)(y+3)
\][/tex]
4. Factorizing \(3 y^2 + 18 y + 27\):
[tex]\[
3 y^2 + 18 y + 27 = 3 (y^2 + 6y + 9) = 3 (y+3)^2
\][/tex]
Now, the expression is:
[tex]\[
\frac{2 (y-5)(y+2)}{4 (y+3)} \times \frac{3 (y+3)^2}{(y+2)(y+3)}
\][/tex]
### Step 3: Simplify the Expression
Next, we simplify the expression by canceling out common factors in the numerator and the denominator:
[tex]\[
\frac{2 (y-5)(y+2)}{4 (y+3)} \times \frac{3 (y+3)^2}{(y+2)(y+3)} = \frac{2 \cancel{(y-5)(y+2)}}{4 \cancel{(y+3)}} \times \frac{3 \cancel{(y+3)^2}}{\cancel{(y+2)} \cancel{(y+3)}}
\][/tex]
After canceling, we have:
[tex]\[
\frac{2 (y-5)}{4} \times \frac{3 (y+3)}{y+3}
\][/tex]
Since \(y+3\) cancels out, we get:
[tex]\[
\frac{2 (y-5)}{4} \times 3 = \frac{3}{2} (y-5)
\][/tex]
So, the simplified result is:
[tex]\[
\frac{3}{2} (y-5)
\][/tex]
### Step 4: Select the Correct Answer
On inspecting the given multiple choices, the expression simplifies to:
[tex]\[
\boxed{\frac{3(y-5)}{2}}
\][/tex]
