Problem 95 - College Algebra with Modeling and Visualization

Question

Simplify. $$ \frac{x}{x^{2}-4}-\frac{1}{x^{2}+4 x+4} $$

Step-by-Step Solution

Answer

\(\frac{x^2 + x + 2}{(x-2)(x+2)^2}\)

Step 1: Identify the Common Denominator

First, observe the denominators: \(x^2 - 4\) and \(x^2 + 4x + 4\). Notice that \(x^2 - 4\) can be factored as \((x-2)(x+2)\), and \(x^2 + 4x + 4\) can be factored as \((x+2)^2\). The least common denominator (LCD) is \((x-2)(x+2)^2\).

Step 2: Rewrite Each Fraction with the LCD

Convert each fraction to have the LCD: 1. For \(\frac{x}{x^2 - 4}\), multiply both numerator and denominator by \(x+2\) to get \(\frac{x(x+2)}{(x-2)(x+2)^2}\).2. For \(\frac{1}{x^2 + 4x + 4}\), multiply both numerator and denominator by \(x-2\) to get \(\frac{1(x-2)}{(x-2)(x+2)^2}\).

Step 3: Combine the Fractions

Now that both fractions have a common denominator, combine them:\[\frac{x(x+2) - (x-2)}{(x-2)(x+2)^2}\]

Step 4: Simplify the Numerator

Simplify the expression in the numerator:First, distribute and combine like terms:\[x(x+2) = x^2 + 2x\]Subtract the second part:\[x^2 + 2x - (x - 2) = x^2 + 2x - x + 2 = x^2 + x + 2\]

Step 5: Simplify the Expression

Finally, express the simplified fraction:\[\frac{x^2 + x + 2}{(x-2)(x+2)^2}\] Check for any further simplification, but in this case, the expression is already in its simplest form.

Key Concepts

Factoring

Factoring is a fundamental skill in algebra that makes manipulating mathematical expressions simpler and more approachable. The process of factoring involves expressing a polynomial as the product of its factors, essentially reversing the distributive property. Recognizing different forms of polynomials is the first step. Let's look at two expressions:

• The expression \(x^2 - 4\) is a 'difference of squares' and can be factored into \((x-2)(x+2)\). This identity is crucial because it shows how two squares' difference can be broken down into two linear factors.
• The expression \(x^2 + 4x + 4\) is a 'perfect square trinomial'. It factors into \((x+2)^2\). Understanding this concept helps simplify expressions by recognizing when you have squared terms in trinomials.

Mastering these core factoring identities and recognizing patterns are key to simplifying more complex algebraic fractions. As fractions with polynomials are common in algebra, factoring helps in determining relationships between numerators and denominators.

Least Common Denominator

The Least Common Denominator (LCD) is an essential concept when dealing with algebraic fractions, acting as a partner to help you combine different fractions into a single entity. To find the LCD, you must identify and use the greatest number of each factor present in any of the original denominators. Here’s how it applies to our problem:

• We have denominators \((x-2)(x+2)\) and \((x+2)^2\).
• The LCD is \((x-2)(x+2)^2\) because it incorporates each factor the maximum times it appears in any denominator. In this case, "\((x+2)\)" appears squared in one of the denominators.

The LCD is crucial as it allows you to rewrite expressions with a uniform base, facilitating their addition or subtraction. By ensuring all fractions share this common denominator, students can then focus on simplifying the numerators, making tasks much easier to manage.

Polynomial Simplification

Polynomial simplification involves reducing expressions to their most basic form, making them easier to work with. Once the fractions share a common denominator, as shown through the LCD, you can simplify the expression. To simplify effectively, follow these steps:

• Combine the numerators as one, ensuring any addition or subtraction is applied properly: \[x(x+2) - (x-2)\].
• Distribute and organize the like terms: By distributing the terms, you perform operations such as \[x \cdot x + 2x - x + 2\], which results in \(x^2 + x + 2\).
• Always check if the simplified numerator can factor further or be reduced with the denominator, although sometimes, as in our case, it may already be in its simplest form.

Being able to simplify polynomial expressions improves algebraic literacy, enhances the ability to see nuanced relationships between polynomial components, and streamlines error checking by working with the simplest possible terms.