Problem 39 - College Algebra and Calculus: An Applied Approach

Question

Perform the indicated operations and simplify.\(\frac{r}{r-1} \cdot \frac{r^{2}-1}{r^{2}}\)

Step-by-Step Solution

Answer

The simplified term is \( r^{2} + r \)

Step 1: Factorization

The numerator in the second fraction is a difference of squares, namely, \( r^{2}-1 \), which can be factored to \( (r - 1)(r + 1) \). Therefore, the expression now looks like this: \(\frac{r}{r-1} \cdot \frac{(r - 1)(r + 1)}{r^{2}} \)

Step 2: Simplify The Expression

Next step is to cancel out common terms from the numerator and denominator. It can be noticed that the term \( r - 1 \) in the numerator and denominator cancels out, leaving the expression as: \( r \cdot \frac{r + 1}{r^{2}} \)

Step 3: Distribute the r Term

The last step is to distribute \( r \) into the fraction. This results in \( r*r + r*1 \) which simplifies to \( r^{2} + r \). This is the final simplified form of the expression.

Key Concepts

Difference of Squares

In algebra, the term 'Difference of Squares' refers to a specific pattern used in factoring expressions. This pattern occurs in expressions like \( a^2 - b^2 \), which represents the difference between two perfect squares. It's a useful technique because it allows us to rewrite these expressions into a product of simpler binomials.

The formula for factoring a difference of squares is:

• \( a^2 - b^2 = (a - b)(a + b) \)

For example, in the given expression \( r^2 - 1 \), it can be observed that it fits the difference of squares pattern where \( a = r \) and \( b = 1 \). Thus, it can be factored easily into \( (r - 1)(r + 1) \). Recognizing this pattern is key in simplifying complex algebraic fractions as it allows you to break down and work with simpler components.

Factoring

Factoring is a fundamental process in algebra involving the breaking down of expressions into simpler multipliers or factors. This process is invaluable, especially when simplifying complex algebraic fractions.

The initial step in tackling the original problem is recognizing which parts of the expression need to be factored. Here, the numerator of the second fraction, \( r^2 - 1 \), is identified as a candidate for factoring due to its difference of squares configuration, which resolves into \( (r - 1)(r + 1) \).When working with algebraic fractions, always keep an eye out for common factoring techniques:

• Greatest Common Factor (GCF): Look for and factor out the largest common factor in an expression.
• Difference of Squares: Factor expressions of the form \( a^2 - b^2 \).
• Trinomials: Factor quadratic trinomials into binomials when possible.

Using these methods, especially with complex fractions, helps streamline the simplification and solution process.

Simplification Steps

Simplifying algebraic fractions involves systematically reducing the expression to its simplest form by canceling out common factors in the numerator and the denominator.

In the given exercise, after factorizing \( r^2 - 1 \) into \( (r - 1)(r + 1) \), you are left with the expression:\[\frac{r}{r-1} \cdot \frac{(r - 1)(r + 1)}{r^2}.\]The next step involves simplification by canceling the common term \( r - 1 \) found in the numerator of the second fraction and the denominator of the first fraction.Here’s a concise plan to simplify:

• Identify common terms in the numerator and denominator across the fractions.
• Cancel out these common terms using the property \( \frac{a}{a} = 1 \).
• Distribute remaining terms if needed, as with the multiplication \( r \cdot \frac{r + 1}{r^2} \).

Following these steps leaves the expression as \( r^2 + r \), which is now its simplest form. Simplification is crucial for solving algebra problems accurately and efficiently, ensuring a streamlined path to the answer.