Finding The Quotient Of Polynomial Expressions

In mathematics, the quotient represents the result of a division operation. To truly grasp the concept of a quotient, we need to delve into the process of polynomial division. This article aims to provide a comprehensive explanation of how to determine the quotient, using a specific example to illustrate the steps involved. We'll break down the problem, explore the underlying principles, and arrive at the correct solution, ensuring a clear understanding of this fundamental mathematical concept.

Deconstructing the Problem: Diving Polynomial Expressions

Let's tackle the problem head-on. We are given the following expression:

$\frac{2 y^2-6 y-20}{4 y+12} \div \frac{y^2+5 y+6}{3 y^2+18 y+27}$

Our mission is to find the quotient when the first rational expression is divided by the second. To achieve this, we'll embark on a step-by-step journey, employing factorization, simplification, and the crucial principle of "invert and multiply." By meticulously breaking down each stage, we'll transform this seemingly complex expression into a manageable form, ultimately revealing the quotient and solidifying our grasp of polynomial division. The initial step in simplifying this expression involves transforming the division problem into a multiplication problem, which is achieved by inverting the second fraction and changing the division sign to a multiplication sign. This transformation is based on the fundamental principle that dividing by a fraction is the same as multiplying by its reciprocal. By applying this rule, we set the stage for further simplification and ultimately determine the quotient of the given polynomial expressions.

Step 1: Transform Division into Multiplication

Remember the golden rule of dividing fractions: dividing by a fraction is the same as multiplying by its reciprocal. So, we flip the second fraction and change the division to multiplication:

$\frac{2 y^2-6 y-20}{4 y+12} \times \frac{3 y^2+18 y+27}{y^2+5 y+6}$

This crucial step transforms our division problem into a multiplication problem, paving the way for simplification through factorization. By inverting the second fraction and changing the operation to multiplication, we set the stage for applying factoring techniques to both the numerators and denominators. This transformation is not merely a trick; it's a fundamental principle in fraction manipulation, allowing us to work with multiplication, which is often more straightforward than division, especially when dealing with algebraic expressions. The act of inverting the fraction essentially provides us with the multiplicative inverse, which, when multiplied by the original fraction, results in unity. This concept is pivotal in solving equations and simplifying expressions in algebra.

Step 2: Factorize Each Polynomial

Now comes the heart of the problem: factoring. We need to break down each polynomial into its simplest factors. This process involves identifying common factors, recognizing patterns like the difference of squares or perfect square trinomials, and skillfully applying the distributive property in reverse. Factoring is not just a mechanical process; it's a crucial skill that unlocks the underlying structure of algebraic expressions, allowing us to simplify, solve equations, and gain deeper insights into mathematical relationships. In this specific problem, factoring will enable us to identify common factors between the numerator and denominator, which can then be canceled out, leading to the simplified form of the expression and ultimately revealing the quotient.

Factor the first numerator:

$2 y^2-6 y-20 = 2(y^2 - 3y - 10) = 2(y-5)(y+2)$

First, we factor out the common factor of 2. Then, we factor the quadratic expression. We look for two numbers that multiply to -10 and add to -3. Those numbers are -5 and 2.

Factor the first denominator:

$4y + 12 = 4(y+3)$

This is a straightforward factorization, pulling out the common factor of 4.

Factor the second numerator:

$3 y^2+18 y+27 = 3(y^2 + 6y + 9) = 3(y+3)(y+3) = 3(y+3)^2$

Again, we start by factoring out the common factor of 3. The remaining quadratic is a perfect square trinomial, which factors neatly into $(y+3)^2$.

Factor the second denominator:

$y^2+5 y+6 = (y+2)(y+3)$

Here, we need two numbers that multiply to 6 and add to 5. Those numbers are 2 and 3.

Step 3: Rewrite the Expression with Factored Polynomials

Now, we substitute the factored forms back into our expression:

$\frac{2(y-5)(y+2)}{4(y+3)} \times \frac{3(y+3)^2}{(y+2)(y+3)}$

This step is crucial as it visually presents the factored forms of the polynomials, making it easier to identify common factors that can be canceled out. By rewriting the expression in this manner, we create a clear roadmap for simplification, highlighting the elements that contribute to the final quotient. The factored form not only facilitates cancellation but also provides valuable insights into the roots and behavior of the polynomials, which are essential in various mathematical applications, such as solving equations and graphing functions. This step underscores the importance of mastering factorization techniques as a fundamental skill in algebra.

Step 4: Simplify by Canceling Common Factors

This is where the magic happens! We can cancel out factors that appear in both the numerator and the denominator. Remember, canceling is essentially dividing both the numerator and denominator by the same factor, which doesn't change the value of the expression. The ability to identify and cancel common factors is a cornerstone of simplifying algebraic expressions, leading to more manageable forms and revealing the underlying relationships between the variables. This step not only reduces the complexity of the expression but also provides a clearer picture of its behavior, making it easier to analyze and interpret in various mathematical contexts.

• We can cancel a $(y+2)$ from the numerator and denominator.
• We can cancel a $(y+3)$ from the numerator and denominator.
• We can simplify the constants: $\frac{2}{4} \times 3 = \frac{1}{2} \times 3 = \frac{3}{2}$

After canceling common factors, our expression looks like this:

$\frac{(y-5)}{2} \times \frac{3(y+3)}{ (y+3)}$

$\frac{(y-5)}{2} \times \frac{3}{1}$

Step 5: Multiply Remaining Terms

Finally, we multiply the remaining terms in the numerator and the denominator:

$\frac{3(y-5)}{2}$

This final multiplication consolidates the simplified factors into a single expression, representing the quotient of the original division problem. The result is a concise and easily interpretable form, showcasing the power of factorization and simplification techniques in algebraic manipulation. This step not only provides the answer to the specific problem but also reinforces the fundamental principles of fraction multiplication and the importance of clear and organized mathematical procedures.

The Answer

The quotient is $\frac{3(y-5)}{2}$, which corresponds to option B.

In conclusion, finding the quotient of polynomial expressions involves a series of steps: transforming division into multiplication, factoring each polynomial, canceling common factors, and finally, multiplying the remaining terms. By mastering these techniques, you can confidently tackle complex algebraic problems and gain a deeper understanding of mathematical relationships. This step-by-step approach not only leads to the correct answer but also cultivates a systematic problem-solving mindset, which is invaluable in various mathematical and scientific disciplines. The ability to break down complex problems into manageable steps and apply appropriate techniques is a hallmark of mathematical proficiency, empowering individuals to tackle challenges with confidence and precision.