- Rewrite the division as multiplication by the reciprocal.
- Factor all polynomials in the numerators and denominators: \(2(y-5)(y+2)\), \(4(y+3)\), \(3(y+3)^2\), and \((y+2)(y+3)\).
- Cancel all common factors, leaving \(\frac{6(y-5)}{4}\).
- Simplify the constant fraction to get \(\frac{3(y-5)}{2}\).
- Final answer: \(\boxed{\frac{3(y-5)}{2}}\)
Explanation
- Rewriting and Factoring the Expressions
To simplify the quotient \(\frac{2y^2-6y-20}{4y+12} \div \frac{y^2+5y+6}{3y^2+18y+27}\), we start by rewriting the division as multiplication by the reciprocal of the second fraction:
\[ \frac{2y^2-6y-20}{4y+12} \cdot \frac{3y^2+18y+27}{y^2+5y+6} \]
Next, we factor each quadratic and linear expression:
- Numerator of the first fraction: \(2y^2-6y-20 = 2(y^2-3y-10) = 2(y-5)(y+2)\).
- Denominator of the first fraction: \(4y+12 = 4(y+3)\).
- Numerator of the second fraction: \(3y^2+18y+27 = 3(y^2+6y+9) = 3(y+3)^2\).
- Denominator of the second fraction: \(y^2+5y+6 = (y+2)(y+3)\).
- Simplification by Cancellation
Substituting the factored expressions back into our multiplication setup:
\[ \frac{2(y-5)(y+2)}{4(y+3)} \cdot \frac{3(y+3)^2}{(y+2)(y+3)} \]
Now, we simplify the expression by canceling common factors in the numerator and denominator:
- The factor \((y+2)\) appears in the first numerator and second denominator, so it cancels out.
- The factor \((y+3)\) appears in the first denominator and twice in the second numerator (since \((y+3)^2 = (y+3)(y+3)\)). One \((y+3)\) cancels with the one in the first denominator, and the second one cancels with the one in the second denominator.
After cancellation, we have:
\[ \frac{2(y-5) \cdot 3}{4} = \frac{6(y-5)}{4} = \frac{3(y-5)}{2} \]
The simplified expression is \(\frac{3(y-5)}{2}\). Comparing this result with the given choices, it matches the second option.
Examples
Simplifying rational expressions is essential in fields like physics and engineering, where complex formulas often describe motion or circuit behavior. For instance, when analyzing the gain of an electronic filter, you might arrive at a complex ratio of polynomials. Reducing these expressions to their simplest form helps engineers understand the system's response more intuitively and perform faster calculations without losing accuracy. This process is effectively the same as simplifying the math problem above, allowing for a clearer understanding of the underlying variables.