Problem 32 - Introductory Algebra for College Students

Question

Multiply as indicated.$ $$\frac{x^{2}-y^{2}}{x+y} \cdot \frac{x+2 y}{2 x^{2}-x y-y^{2}}$$

Step-by-Step Solution

Answer

The simplified version of the given expression is \( \frac{x+2y}{2x+y} \).

Step 1: Rewrite the equation

Let's rewrite the equation without the multiplication sign: \( \frac{x^{2}-y^{2}}{x+y} \times \frac{x+2 y}{2 x^{2}-x y-y^{2}} \) to \( \frac{x^{2}-y^{2}}{x+y} \cdot \frac{x+2 y}{2 x^{2}-x y-y^{2}} \).

Step 2: Factorize the expressions

Applying a factorization technique, difference squares on \( x^{2}-y^{2} \), it becomes \( (x-y)(x+y) \). Similarly, the denominator \( 2 x^{2}-x y-y^{2} \) can be written as \( (x- y)(2x+y) \). Thus our expression now becomes \( \frac{(x-y)(x+y)}{x+y} \cdot \frac{(x+2 y)}{(x-y)(2x+y)} \)

Step 3: Simplify the expressions

It's easy to see that, \( x+y \) in the numerator and denominator and \( x-y \) can be cancelled out. So, the overall expression simplifies to \( \frac{x+2y}{2x+y} \)

Key Concepts

Difference of Squares

The difference of squares is a fundamental algebraic concept that comes into play when we encounter expressions like \( x^2 - y^2 \). It is predicated on the identity \( a^2 - b^2 = (a + b)(a - b) \). This identity implies that when we have a subtraction between two squared terms, we can factor it into the product of two binomials. For example, in the exercise \( x^2 - y^2 \), we recognize this as a difference of squares and thus factor it as \( (x + y)(x - y) \).

The ability to factor these expressions is not just a neat algebraic trick; it is a powerful tool that simplifies complex algebraic expressions and is essential for solving equations and understanding higher-level mathematics. It can also reveal symmetries and help in graphing parabolic equations. Factoring differences of squares allows for the cancellation of terms when in fraction form, as seen in the step-by-step solution provided.

Simplifying Algebraic Expressions

Why Simplification Matters

When working with algebraic expressions, simplification is the process of reducing complexity, making expressions more understandable and easier to work with. Simplification can include factoring expressions, cancelling out like terms, or performing arithmetic operations.

In the original exercise, after applying the difference of squares, we obtained \( \frac{(x-y)(x+y)}{x+y} \cdot \frac{(x+2y)}{(x-y)(2x+y)} \). Here, simplification includes recognizing that \( x+y \) in the numerator and denominator, as well as \( x-y \), cancel each other out because they are common factors to both numerator and denominator. This cancellation is fundamental to simplification, helping us reach the more manageable form \( \frac{x+2y}{2x+y} \).

This principle is not just limited to the exercise in question but is a critical skill in algebra that enables students to progress to more complex mathematics. Simplifying expressions helps in solving equations, analyzing functions, and calculating derivatives in calculus. It's all about making the problem easier to digest and solve.

Multiplying Rational Expressions

Understanding Rational Expressions

Rational expressions are fractions that involve polynomials. When multiplying rational expressions, the process is similar to multiplying numerical fractions: multiply the numerators together and multiply the denominators together. However, before doing this, it’s beneficial to factor the polynomials to see if any terms can be canceled to simplify the expression.

In the given problem, we factored the polynomials to identify and cancel out the common terms. This step is crucial, as it can significantly simplify the multiplication process. After canceling, we multiply what’s left of the numerators and denominators to get the final simplified expression.

Multiplying rational expressions requires a strong understanding of factoring and simplifying. Students should feel comfortable with these processes to handle rational expressions efficiently, as it comes in handy not just in algebra but also in calculus when dealing with complex rational functions.