Problem 59 - College Algebra and Calculus: An Applied Approach

Question

Perform the indicated operations and simplify.\(\frac{2}{x^{2}-4}-\frac{1}{x^{2}-3 x+2}\)

Step-by-Step Solution

Answer

\The simplified form of the given expression is \(\frac{x-4}{(x-2)(x+2)(x-1)}\)

Step 1: Factorize the Denominators

Rewrite the denominators of the fractions as the product of their factors. In this case, \(x^{2}-4\) can be written as \((x-2)(x+2)\), and \(x^{2}-3x+2\) can be written as \((x-2)(x-1)\). So the expressions become \(\frac{2}{(x-2)(x+2)}\) and \(\frac{1}{(x-2)(x-1)}\) respectively.

Step 2: Find the Common Denominator

The common denominator of two fractions is found by taking the product of the different factors present in their denominators. In this case, the common denominator is \((x-2)(x+2)(x-1)\).

Step 3: Multiply Numerator and Denominator by Missing Factor

For the first fraction, its denominator is missing the factor \((x-1)\). Therefore, multiply its numerator and denominator by \((x-1)\), and for the second fraction, its denominator is missing \((x+2)\). Therefore, multiply its numerator and denominator by \((x+2)\). So we get \(\frac{2(x-1)}{(x-2)(x+2)(x-1)}\) and \(\frac{(x+2)}{(x-2)(x+2)(x-1)}\).

Step 4: Simplify Expression

Subtract the second fraction from the first: \(\frac{2(x-1)-(x+2)}{(x-2)(x+2)(x-1)}\). Perform the subtraction in the numerator to get: \(\frac{(2x-2)-(x+2)}{(x-2)(x+2)(x-1)}\) = \(\frac{x-4}{(x-2)(x+2)(x-1)}\). This is the simplified form of the given expression.

Key Concepts

Factoring Polynomials

Factoring polynomials is like breaking down a number into its prime factors. You take a polynomial and express it as a product of simpler polynomials. This process is essential because it often makes subsequent mathematics operations easier. For example, when you have the polynomial \(x^2 - 4\), you can recognize it as a difference of squares, which factors into \((x-2)(x+2)\). Similarly, the quadratic \(x^2 - 3x + 2\) can be factored by finding two numbers that multiply to 2 and add up to -3, which are -1 and -2. Hence, \(x^2 - 3x + 2\) factors into \((x-2)(x-1)\). This step is pivotal for simplifying complex algebraic expressions.

Common Denominator

When working with fractions that have different denominators, finding a common denominator makes it possible to combine the fractions. A common denominator is simply a shared multiple of the original denominators. In rational expressions, the least common denominator (LCD) is preferred as it simplifies calculations. In our problem, the denominators are \((x-2)(x+2)\) and \((x-2)(x-1)\). To find the common denominator, combine the unique factors: \((x-2)(x+2)(x-1)\). With this common denominator, you can easily add or subtract the fractions.

Simplifying Algebraic Expressions

Simplifying algebraic expressions involves reducing complexity while retaining equivalence. After combining fractions through a common denominator, the next step is simplifying the numerator. Use algebraic manipulation, such as distributing and combining like terms, to simplify. For instance, in the expression \(2(x-1) - (x+2)\), you simplify it by multiplying and then distributing, leading to \(2x - 2 - x - 2 = x - 4\). Simplifying reduces the expression to its most basic form, making it easier to understand or further manipulate.

Fraction Operations

Fraction operations involve adding, subtracting, multiplying, and dividing fractions. Each operation follows specific rules, especially when variable expressions are involved. For arithmetic on fractions with different denominators, ensure to first find a common denominator. As in the problem, subtraction requires transforming fractions with different denominators to equivalent fractions with the same denominator, \((x-2)(x+2)(x-1)\). This harmonization allows the numerators to be subtracted directly, thus simplifying the complex algebraic fraction into a single fraction form, \(\frac{x-4}{(x-2)(x+2)(x-1)}\).