Problem 14 - College Algebra

Question

For the following exercises, 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

The first step is to factor the denominator of the given fraction. The denominator is \(x^2-25\), which is a difference of squares. It can be factored as \((x+5)(x-5)\). So, the expression becomes \(\frac{10x}{(x+5)(x-5)}\).

Step 2: Set Up Partial Fraction Form

Since the denominator consists of two nonrepeating linear factors \((x+5)\) and \((x-5)\), we can express the partial fraction decomposition as: \(\frac{A}{x+5} + \frac{B}{x-5}\), where \(A\) and \(B\) are constants we need to determine.

Step 3: Clear the Denominator

Multiply the entire equation by the common denominator \((x+5)(x-5)\) to eliminate the denominators. This yields: \[ 10x = A(x-5) + B(x+5). \]

Step 4: Expand and Simplify

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

Step 5: Solve for the Constants

Since the equation \(10x = (A + B)x + (5B - 5A)\) must hold for all values of \(x\), we equate the coefficients of like terms. For the \(x\)-terms: \[ A + B = 10, \]and for the constant terms: \[ 5B - 5A = 0. \] Simplifying the second equation, we have: \[ B = A. \] Substitute \(B = A\) into the first equation: \[ A + A = 10 \]\[ 2A = 10 \] \[ A = 5. \] Since \(B = A\), \(B = 5\).

Step 6: Write the Partial Fraction Decomposition

Substitute \(A = 5\) and \(B = 5\) back into the partial fraction form: \[ \frac{10x}{x^2 - 25} = \frac{5}{x+5} + \frac{5}{x-5}. \] This is the partial fraction decomposition of the given expression.

Key Concepts

Nonrepeating Linear Factors

When dealing with partial fraction decomposition, particularly with nonrepeating linear factors, the goal is to break down a complex fraction into simpler, more manageable components. Nonrepeating linear factors are linear expressions in the denominator that do not repeat, like

• \( (x+5) \)
• \( (x-5) \).

In our exercise, after factoring the denominator using the difference of squares rule, we are left with these two distinct linear factors. Each factor corresponds to a separate term in the decomposition.
For a denominator with linear factors, we set up the partial fraction decomposition by assigning a separate constant to each factor. So, for

• \( \frac{10x}{(x+5)(x-5)} \)

We write:

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

This transforms a single complex expression into a sum of simpler fractions, making it easier to integrate or differentiate if needed.

Difference of Squares

The difference of squares is a special factoring pattern that occurs when you have two squared terms separated by a subtraction sign, such as \( x^2 - 25 \). This expression can be factored into

• \( (x+5)(x-5) \).

The pattern can be generally expressed as:

• \( a^2 - b^2 = (a+b)(a-b) \).

Recognizing the difference of squares allows you to simplify complex fractions and identify potential non-repeating linear factors, which are crucial for setting up partial fraction decomposition. It is a fundamental algebraic tool that aids in breaking down expressions, facilitating easier manipulation and evaluation of algebraic fractions.

Algebraic Fractions

Algebraic fractions are similar to regular fractions but contain polynomials in the numerator, the denominator, or both. Understanding how to manipulate these expressions is a key aspect of algebra. The partial fraction decomposition is a method used to express algebraic fractions, especially those with polynomial denominators, as a sum of simpler fractions.
In the exercise example,

• \( \frac{10x}{x^2 - 25} \)

becomes much easier to work with once decomposed into

• \( \frac{5}{x+5} + \frac{5}{x-5} \).

This decomposition leverages the factored form of the denominator and allocates a constant term to each factor. This approach is extremely helpful for solving integrals, comparing expressions, and simplifying calculations in calculus and advanced mathematics.