Problem 20 - Elementary and Intermediate Algebra

Question

The LCD for \(\frac{2 x+1}{x^{2}+5 x+6}\) and \(\frac{3 x}{x^{2}-4}\) is $$\mathrm{LCD}=(x+2)(x+3)(x-2)$$ If we want to subtract these rational expressions, what form of 1 should be used a. to build \(\frac{2 x+1}{x^{2}+5 x+6} ?\) b. to build \(\frac{3 x}{x^{2}-4} ?\)

Step-by-Step Solution

Answer

a. Use \(\frac{x-2}{x-2}\) for \(\frac{2x+1}{x^2+5x+6}\). b. Use \(\frac{x+3}{x+3}\) for \(\frac{3x}{x^2-4}\).

Step 1: Understand the Problem

We are given two rational expressions, \(\frac{2x+1}{x^2+5x+6}\) and \(\frac{3x}{x^2-4}\), and need to subtract them. To do this, we must convert both to have a common denominator, the Least Common Denominator (LCD).

Step 2: Factor Denominators

First, factor the denominators of the given expressions. The denominator \(x^2+5x+6\) factors to \((x+2)(x+3)\), and \(x^2-4\) factors to \((x-2)(x+2)\).

Step 3: Verify the LCD

The LCD for the two rational expressions needs to include each of the factors from the individual denominators. Thus, the LCD is \((x+2)(x+3)(x-2)\). This is given in the problem statement as the correct LCD.

Step 4: Adjust First Expression to LCD

The first expression, \(\frac{2x+1}{(x+2)(x+3)}\), needs to adjust its denominator to match the LCD. It is missing the factor \((x-2)\). Multiply the numerator and denominator by \((x-2)\): \[ \frac{2x+1}{(x+2)(x+3)} \times \frac{(x-2)}{(x-2)} = \frac{(2x+1)(x-2)}{(x+2)(x+3)(x-2)} \]

Step 5: Adjust Second Expression to LCD

The second expression, \(\frac{3x}{(x-2)(x+2)}\), needs to adjust its denominator to match the LCD. It is missing the factor \((x+3)\). Multiply the numerator and denominator by \((x+3)\): \[ \frac{3x}{(x-2)(x+2)} \times \frac{(x+3)}{(x+3)} = \frac{3x(x+3)}{(x-2)(x+2)(x+3)} \]

Key Concepts

Factoring Polynomials

When dealing with rational expressions, factoring polynomials is a crucial skill. Factoring involves rewriting a polynomial as a product of simpler polynomials. This makes it easier to simplify expressions and find solutions.

Given the exercise, we had two denominators to factor:

• The first denominator, \(x^2 + 5x + 6\), factors to \((x+2)(x+3)\).
• The second denominator, \(x^2 - 4\), factors to \((x-2)(x+2)\). This is known as a difference of squares.

Understanding how to factor polynomials can simplify complex algebraic expressions. Always look for opportunities to factor expressions. It greatly simplifies the handling of rational equations.

Subtracting Rational Expressions

Subtracting rational expressions requires a common denominator, just like with numerical fractions. If the denominators differ, you'll need to find a common denominator to simplify the computation. In the exercise, the expressions were:

• \(\frac{2x+1}{(x+2)(x+3)}\)
• \(\frac{3x}{(x-2)(x+2)}\)

Subtracting them directly would have been complex. Instead, by using the Least Common Denominator (LCD), we can easily perform the subtraction. Begin by making sure both expressions share the LCD, then subtract by applying the distributive property if necessary.

Subtracting means distributing any minus sign through all terms in the numerators properly, ensuring no simple mistakes are made.

Common Denominator

Finding the common denominator is key to combining rational expressions. The LCD is the smallest expression that both denominators can divide without a remainder. In this exercise, the LCD was found by multiplying each unique factor of the denominators:

• \( (x+2)(x+3)(x-2) \)

This includes all factors, ensuring compatibility between the expressions. To apply the LCD:

1. Adjust the numerators by multiplying them by the missing factors from the LCD.
2. Ensure both numerators have the same denominators before performing operations like addition or subtraction.

Having the common denominator makes any arithmetic operation on rational expressions straightforward and reduces the risk of algebraic errors.