Problem 43 - Algebra A Combined Function

Question

Perform each indicated operation. Simplify if possible. \(\frac{x}{x^{2}-1}-\frac{2}{x^{2}-2 x+1}\)

Step-by-Step Solution

Answer

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

Step 1: Factor Denominators

First, we need to try and factor each of the denominators in the expressions. The first denominator is \(x^2 - 1\), which is a difference of squares and can be factored as \((x+1)(x-1)\). The second denominator is \(x^2 - 2x + 1\), which is a perfect square trinomial and factors as \((x-1)^2\).

Step 2: Find Least Common Denominator

To combine the fractions, we need a common denominator. The least common denominator (LCD) for \((x+1)(x-1)\) and \((x-1)^2\) is \((x+1)(x-1)^2\).

Step 3: Rewrite each Fraction with the LCD

We need to adjust each fraction to have the LCD as the denominator. For the first fraction, \(\frac{x}{(x+1)(x-1)}\), multiply both the numerator and the denominator by \(x-1\) to get \(\frac{x(x-1)}{(x+1)(x-1)^2}\). For the second fraction, \(\frac{2}{(x-1)^2}\), multiply both the numerator and the denominator by \(x+1\) to get \(\frac{2(x+1)}{(x+1)(x-1)^2}\).

Step 4: Perform the Subtraction

Subtract the second fraction from the first with the common denominator: \(\frac{x(x-1) - 2(x+1)}{(x+1)(x-1)^2}\). Expand and combine like terms in the numerator: \(x^2 - x - 2x - 2 = x^2 - 3x - 2\).

Step 5: Simplify the Expression

The resulting expression is \(\frac{x^2 - 3x - 2}{(x+1)(x-1)^2}\). Check if the numerator can be factored further. It factors as \((x-2)(x-1)\). Thus we have \(\frac{(x-2)(x-1)}{(x+1)(x-1)^2}\).

Step 6: Cancel Common Factors

Cancel the common factor \((x-1)\) from the numerator and the denominator: \(\frac{x-2}{(x+1)(x-1)}\). This is the simplified form.

Key Concepts

Factoring Polynomials

When dealing with algebraic expressions, factoring polynomials is a critical skill. It involves breaking down a polynomial into simpler components called factors. These factors, when multiplied together, give back the original polynomial.

For instance, in the exercise, the first polynomial in the denominator is \(x^2 - 1\). This is a "difference of squares" because it can be expressed as \(a^2 - b^2\), which factors into \((a + b)(a - b)\). Here, \(x^2 - 1\) becomes \((x+1)(x-1)\).

• Recognizing patterns like difference of squares can simplify the factoring process greatly.
• Another common pattern is "perfect square trinomials," such as \(x^2 - 2x + 1\), which factors to \((x-1)^2\).

Understanding how to factor polynomials helps manage complex expressions and is crucial for simplifying fractions located within algebraic equations.

Common Denominators

To perform operations like subtraction or addition on fractions, a common denominator is essential. This means converting each fraction so their denominators are identical, allowing the numerators to be combined directly.

In the example problem, the denominators \((x+1)(x-1)\) and \((x-1)^2\) need a least common denominator (LCD). The LCD is the smallest expression that both denominators divide into evenly.

• This ensures the fractions can be easily subtracted or added.
• In this case, the LCD derived is \((x+1)(x-1)^2\).

To achieve this, each fraction's numerator and denominator are adjusted accordingly. The first fraction is modified by multiplying by \(x-1\), and the second by \(x+1\), ensuring they both share this common denominator.

Simplifying Expressions

Simplifying expressions is the process of making an algebraic expression as basic as possible without changing its value. This involves combining like terms or canceling common factors.

In the solved problem, after combining fractions using a common denominator, the numerator becomes \(x^2 - 3x - 2\). This can be factored into \((x-2)(x-1)\), simplifying further.

• Simplification often requires factoring and reducing by canceling identical terms in the numerator and denominator, as seen by removing \((x-1)\) from both.
• Always check if the expression can be simplified further to ensure clarity and ease in handling future operations.

This step-by-step method to simplification not only simplifies the algebraic expression but also makes solving equations much more manageable and less error-prone.