Problem 70 - Intermediate Algebra

Question

Perform the operations and simplify the result when possible. See Example \(8 .\) $$\frac{x}{x^{2}-4}-\frac{x}{x+2}+\frac{2}{x}$$

Step-by-Step Solution

Answer

Combine the fractions to get \(\frac{2(x^2 + x - 4)}{x(x - 2)(x + 2)}\) as the simplified result.

Step 1: Identify Common Denominator

To combine the fractions, we need a common denominator. The denominators are \(x^2 - 4\), \(x + 2\), and \(x\). Rewrite \(x^2 - 4\) as \((x - 2)(x + 2)\). The common denominator is \(x(x - 2)(x + 2)\).

Step 2: Rewrite Each Fraction

Convert each fraction to have the common denominator: - \(\frac{x}{x^2 - 4}\) becomes \(\frac{x(x)}{x(x - 2)(x + 2)}\)- \(\frac{x}{x+2}\) becomes \(\frac{x(x - 2)}{x(x - 2)(x + 2)}\)- \(\frac{2}{x}\) becomes \(\frac{2(x - 2)(x + 2)}{x(x - 2)(x + 2)}\).

Step 3: Combine the Numerators

Add the numerators of the new fractions: \[ x^2 - x(x - 2) + 2(x - 2)(x + 2) = x^2 - (x^2 - 2x) + 2(x^2 - 4) \] Simplify the expression: \[ x^2 - x^2 + 2x + 2x^2 - 8 = 2x^2 + 2x - 8 \]

Step 4: Simplify the Fraction

The entire fraction is \(\frac{2x^2 + 2x - 8}{x(x - 2)(x + 2)}\). Factor the numerator: \[ 2(x^2 + x - 4) \] Use the quadratic formula to factor \(x^2 + x - 4\) if possible. The factors are not easily obtained, so leave as is. The simplest form is \[ \frac{2(x^2 + x - 4)}{x(x - 2)(x + 2)} \]

Step 5: Final Simplicity Check

No further simplification is possible since the numerator does not share terms with the denominator. Ensure no terms can cancel out.

Key Concepts

Common Denominator

When dealing with rational expressions, finding a common denominator is key to adding or subtracting the fractions. Each rational expression is a fraction with a polynomial in the denominator. To combine these fractions, they must share the same denominator.

To find a common denominator, identify a polynomial that each individual denominator can divide into. In our example, the denominators are \(x^2 - 4\), \(x + 2\), and \(x\). Notice that:

• \(x^2 - 4\) can be rewritten as a product of its factors: \((x - 2)(x + 2)\).
• This revelation is crucial, as it shows that the least common denominator (LCD) is \(x(x - 2)(x + 2)\).

The LCD is a polynomial that incorporates all unique factors from the original denominators, ensuring that when each fraction is rewritten over this shared denominator, the original fractions remain equivalent.

Factoring Polynomials

Factoring polynomials means breaking them down into simpler polynomial factors. This is critical in simplifying expressions or finding a common denominator. Let's delve into how this applies in our problem.

• The expression \(x^2 - 4\) is a difference of squares, which can be written as \((x - 2)(x + 2)\).

Understanding how to factor such polynomials means recognizing patterns, like difference of squares or using the quadratic formula for more complicated polynomials where simple factoring isn't evident.

In this exercise, we attempted to factor the numerator expression, \(2x^2 + 2x - 8\). While using the quadratic formula might help in specific cases, here it was determined we couldn't simplify this further by factoring straightforwardly. However, recognizing when an expression doesn't factor nicely is equally important in streamlining rational expressions.

Simplifying Fractions

Simplifying fractions involves expressing the fraction in its simplest form. This often means reducing the fraction by an equivalent division of the numerator and denominator by their common factors.

• For rational expressions like \(\frac{2(x^2 + x - 4)}{x(x - 2)(x + 2)}\), check if any factors in the numerator are in the denominator.
• If there are no common factors, as in this case, the fraction is already in its simplest form.

Remember, the key step is recognizing common factors between numerator and denominator. Also, check if it is possible to use any special identities or biconditional relationships to further reduce the expression. If not, then ensure every step preserves the equality of the original expression, just in a form that simplifies further arithmetic manipulation.