Problem 14 - College Algebra

Question

Find the decomposition of the partial fraction for the nonrepeating linear factors. \(\frac{10 x}{x^{2}-25}\)

Step-by-Step Solution

Answer

The partial fraction decomposition is \(\frac{5}{x-5} + \frac{5}{x+5}\).

Step 1: Factor the Denominator

To perform partial fraction decomposition, first factor the denominator of the fraction. The given denominator is \(x^2 - 25\), which is a difference of squares. It can be factored as: \((x - 5)(x + 5)\). Thus, we rewrite the fraction as \(\frac{10x}{(x-5)(x+5)}\).

Step 2: Set Up the Partial Fraction Decomposition

Write the expression as a sum of partial fractions. For linear factors \(x-5\) and \(x+5\), we propose:\[\frac{10x}{(x - 5)(x + 5)} = \frac{A}{x - 5} + \frac{B}{x + 5}\]where \(A\) and \(B\) are constants to be determined.

Step 3: Clear the Denominator

Multiply both sides of the equation by the denominator \((x-5)(x+5)\) to eliminate the fractions:\[10x = A(x + 5) + B(x - 5)\]

Step 4: Expand and Simplify

Expand the right side of the equation:\[10x = Ax + 5A + Bx - 5B\]Combine like terms:\[10x = (A + B)x + (5A - 5B)\]

Step 5: Equate Coefficients

Equate the coefficients of corresponding terms on both sides of the equation to form a system of equations:1. \(A + B = 10\) - for the coefficient of \(x\)2. \(5A - 5B = 0\) - for the constant term.

Step 6: Solve the System of Equations

From the second equation, solve for \(A\):\[ 5A = 5B \rightarrow A = B \]Substitute \(A = B\) into the first equation:\[ A + A = 10 \rightarrow 2A = 10 \rightarrow A = 5 \]Thus, \(B = 5\) as well.

Step 7: Write the Partial Fraction Decomposition

Substitute \(A = 5\) and \(B = 5\) back into the partial fractions:\[\frac{10x}{(x - 5)(x + 5)} = \frac{5}{x - 5} + \frac{5}{x + 5}\]

Key Concepts

Nonrepeating Linear Factors

In partial fraction decomposition, nonrepeating linear factors represent terms in the denominator that are linear and do not repeat. This means each factor is of the form \( ax + b \), where \( a \) and \( b \) are constants, and each factor appears only once. For example, in the expression \( \frac{10x}{(x-5)(x+5)} \), the factors \((x-5)\) and \((x+5)\) are nonrepeating linear factors. Since each factor of the denominator is linear and distinct, they don't share the same root or resemble each other.

To set up the partial fraction decomposition, each linear factor becomes the denominator of a separate fraction with an unknown numerator, typically represented by constants like \( A \) and \( B \). This helps in transforming a complex rational expression into simpler, manageable pieces. Here is the setup for decomposition:

• For \((x-5)\), write \( \frac{A}{x-5} \).
• For \((x+5)\), write \( \frac{B}{x+5} \).

By breaking down the expression this way, it becomes easier to manipulate and solve the equations required for determining \( A \) and \( B \). Partial fraction decomposition thereby simplifies integration and other operations involving rational expressions.

Difference of Squares

The concept of difference of squares is a crucial algebraic technique used to factor certain quadratic expressions. A difference of squares follows the pattern \( a^2 - b^2 \), which can always be factored as \( (a-b)(a+b) \). This property is invaluable when dealing with denominators in partial fraction decomposition.

For instance, in the given exercise, the denominator \( x^2 - 25 \) is a difference of squares because it can be rewritten as \( x^2 - 5^2 \). Applying the formula, this expression becomes \( (x-5)(x+5) \). Recognizing and factoring a difference of squares can greatly simplify the computation of partial fractions as seen here. Factoring transforms the problem from a simple polynomial division into a series of straightforward algebraic manipulations.

• Lightens the complexity of the decomposition process.
• Creates separate linear factors that are easier to handle.

Both nonrepeating linear factors derived from a difference of squares allow for a direct setup of partial fractions, leading to a clear path toward finding a solution.

System of Equations

Once you have set up the partial fraction decomposition, determining the unknown constants requires solving a system of equations. A system of equations arises when there are multiple equations with several unknowns. Solving these allows you to find the values of the unknowns, in this case, the constants in your partial fractions.

In the provided problem, once the expression is decomposed into fractions \( \frac{A}{x-5} + \frac{B}{x+5} \), simplifying and combining terms gives us the equation \( 10x = (A + B)x + (5A - 5B) \). By equating the coefficients of corresponding terms (from both sides of the equation), we get the system:

• \( A + B = 10 \)
• \( 5A - 5B = 0 \)

Solving these simultaneous equations will provide the values of \( A \) and \( B \).

To solve the system:

• From the second equation, we find \( A = B \).
• Substitute \( A = B \) into the first equation to find \( A = 5 \). Hence \( B = 5 \).

This process demonstrates how organizing the problem algebraically allows us to find exact values for the unknowns, simplifying the original fraction into its basic components. Thus, using systems of equations is essential for calculating specific elements of a partial fraction decomposition.