- Divide by the reciprocal: rewrite the expression as \(\frac{2 y^2-6 y-20}{4 y+12} \cdot \frac{3 y^2+18 y+27}{y^2+5 y+6}\).
- Factor all parts:
- \(2y^2 - 6y - 20 = 2(y-5)(y+2)\)
- \(4y + 12 = 4(y+3)\)
- \(3y^2 + 18y + 27 = 3(y+3)^2\)
- \(y^2 + 5y + 6 = (y+2)(y+3)\)
- Cancel common factors to obtain \(\frac{6(y-5)}{4}\).
- Simplify the fraction to arrive at the final answer: \(\boxed{\frac{3(y-5)}{2}}\).
Explanation
- Analyzing and Setting Up the Expression
The goal is to simplify the rational expression \(\frac{2 y^2-6 y-20}{4 y+12} \div \frac{y^2+5 y+6}{3 y^2+18 y+27}\). To divide by a fraction, we multiply by its reciprocal:
\[ \frac{2 y^2-6 y-20}{4 y+12} \cdot \frac{3 y^2+18 y+27}{y^2+5 y+6} \]
Our first step is to factor each numerator and denominator completely.
- Factoring the Polynomials
Now, we factor each polynomial individually:
- For \(2y^2 - 6y - 20\), factor out the common factor \(2\): \(2(y^2 - 3y - 10)\). Then, factor the quadratic \(y^2 - 3y - 10\) into \((y-5)(y+2)\). So, \(2y^2 - 6y - 20 = 2(y-5)(y+2)\).
- For \(4y + 12\), factor out the common factor \(4\): \(4(y+3)\).
- For \(3y^2 + 18y + 27\), factor out the common factor \(3\): \(3(y^2 + 6y + 9)\). Recognizing the perfect square trinomial, this becomes \(3(y+3)^2\).
- For \(y^2 + 5y + 6\), factor the quadratic into \((y+2)(y+3)\).
- Simplifying the Expression
Substituting the factored forms back into the multiplication expression:
\[ \frac{2(y-5)(y+2)}{4(y+3)} \cdot \frac{3(y+3)^2}{(y+2)(y+3)} \]
Now, we simplify by canceling common factors in the numerator and denominator:
- The term \((y+2)\) appears in both the numerator and the denominator and cancels out.
- The product \((y+3)(y+3) = (y+3)^2\) in the denominator cancels with \((y+3)^2\) in the numerator.
The expression simplifies to:
\[ \frac{2 \cdot 3 \cdot (y-5)}{4} = \frac{6(y-5)}{4} \]
Dividing the constant coefficients \(6\) and \(4\) by their greatest common divisor, \(2\), we get:
\[ \frac{3(y-5)}{2} \]
This matches the second option provided.
Examples
Rational expressions are used in various fields like engineering and economics to model relationships between variables, such as calculating average speeds or production rates. For instance, if you are designing a system where the efficiency of a machine (represented by a complex fraction) changes depending on input parameters, simplifying that fraction allows you to determine the performance bottleneck quickly. This math exercise helps build the algebraic fluency required to handle these kinds of real-world efficiency models.