Problem 95 - Intermediate Algebra

Question

Perform the operations and simplify the result when possible. Be careful to apply the correct method, because these problems involve addition, subtraction, multiplication, and division of rational expressions. $$\frac{x^{2}+x}{3 x-15} \div \frac{(x+1)^{2}}{6 x-30}$$

Step-by-Step Solution

Answer

The simplified expression is \(\frac{2x}{x + 1}\).

Step 1: Simplify Each Rational Expression in the Division

First, simplify the given rational expressions separately. For the expression \(\frac{x^2 + x}{3x - 15}\), factor common terms. \(x^2 + x\) can be factored into \(x(x + 1)\) and \(3x - 15\) can be factored into \(3(x - 5)\). This gives us the simplified expression \(\frac{x(x + 1)}{3(x - 5)}\). Similarly, for \(\frac{(x + 1)^2}{6x - 30}\), \((x+1)^2\) is already in its simplest form and \(6x - 30\) can be factored into \(6(x - 5)\). Thus, the rational expression simplifies to \(\frac{(x + 1)^2}{6(x - 5)}\).

Step 2: Rewrite the Division as Multiplication by the Reciprocal

When dividing by a fraction, multiply by its reciprocal. Therefore, \(\frac{x(x + 1)}{3(x - 5)} \div \frac{(x + 1)^2}{6(x - 5)}\) becomes \(\frac{x(x + 1)}{3(x - 5)} \cdot \frac{6(x - 5)}{(x + 1)^2}\).

Step 3: Simplify the Expression by Cancelling Common Factors

Next, cancel any common factors between the numerators and denominators. (x - 5) exists in both the numerator and denominator and can be cancelled out. This simplifies the expression to \(\frac{x(x + 1) \cdot 6}{3(x + 1)^2}\). Further simplify by cancelling the \(x + 1\) factor: \(x(x + 1)\) in the numerator and \((x + 1)\) part of \((x + 1)^2\) in the denominator both reduce to \(1\). This leaves us with \(\frac{6x}{3(x + 1)}\).

Step 4: Perform Final Simplification

Finally, simplify \(\frac{6x}{3(x + 1)}\) by factoring out \(3\) from both the numerator and the denominator. This yields \(\frac{2x}{x + 1}\). Thus, the fully simplified expression is \(\frac{2x}{x + 1}\).

Key Concepts

Factoring polynomials

Factoring polynomials is an essential skill needed for working with rational expressions. It's like finding the building blocks of a polynomial expression. The goal is to rewrite a polynomial as a product of its factors, which makes it easier to simplify or perform other operations. Imagine you're trying to break down numbers into their prime factors; the concept is similar! When factoring, look for a common factor as the first step. For example, in the expression:

• \(x^2 + x = x(x + 1)\)
• \(3x - 15 = 3(x - 5)\)

Both are factored by identifying the greatest common factor. Once factored, these expressions reveal simpler components, easing further manipulation or simplification.

Simplifying fractions

Simplifying fractions involves reducing them to their simplest form. This isn't only limited to numerical fractions but also applies to algebraic expressions. A fraction is considered simplified when the numerator and denominator have no common factors other than one. For example, when simplifying the rational expression \(\frac{x^2 + x}{3x - 15}\), factoring it first helps us identify common terms. It becomes \(\frac{x(x + 1)}{3(x - 5)}\). At this point in simplification, always check if any further reduction is possible by cancelling out common terms in the numerator and denominator. This process makes the operations that follow much more straightforward. Remember, the goal is simplicity—get to the most concise form possible while ensuring the mathematical meaning remains unchanged.

Operations with fractions

Operations with fractions, especially with rational expressions, involve addition, subtraction, multiplication, and division. Each of these operations requires specific steps to ensure correctness.

• **Addition and Subtraction**: Requires a common denominator for the fractions involved, much like adding regular fractions.
• **Multiplication**: Multiply across the numerators and the denominators.
• **Division**: Involves multiplying by the reciprocal.

In the given exercise, division is converted to multiplication using the reciprocal. The problem \(\frac{x(x+1)}{3(x-5)} \div \frac{(x+1)^2}{6(x-5)}\) turns into \(\frac{x(x+1)}{3(x-5)} \cdot \frac{6(x-5)}{(x+1)^2}\). This approach enables simplification by cancelling out terms easily.Understanding these operations is vital since one wrong step can lead to incorrect results. Always double-check each step, particularly when rearranging terms or simplifying.