Problem 24 - College Algebra

Question

For the following exercises, find the decomposition of the partial fraction for the repeating linear factors. $$\frac{-24 x-27}{(6 x-7)^{2}}$$

Step-by-Step Solution

Answer

\(\frac{-24x - 27}{(6x-7)^2} = \frac{-4}{6x-7} + \frac{-55}{(6x-7)^2}\).

Step 1: Understand the Problem

We are given a rational function with a repeating linear factor in the denominator, specifically \((6x - 7)^2\). The task is to decompose it into partial fractions. The numerator is \(-24x - 27\).

Step 2: Set Up the Partial Fractions Decomposition

Since the denominator has a repeating linear factor \((6x - 7)^2\), the partial fractions decomposition takes the form: \[ \frac{A}{6x-7} + \frac{B}{(6x-7)^2} \]. Here, \(A\) and \(B\) are constants that need to be determined.

Step 3: Express the Equation

Express the equation by setting the given fraction equal to the sum of partial fractions: \[ \frac{-24x - 27}{(6x-7)^2} = \frac{A}{6x-7} + \frac{B}{(6x-7)^2} \].

Step 4: Eliminate the Denominator

Multiply both sides by \((6x - 7)^2\) to eliminate the denominator: \[ -24x - 27 = A(6x - 7) + B \].

Step 5: Expand and Collect Like Terms

Expand the right-hand side: \[ A(6x - 7) = 6Ax - 7A \] So, \[ -24x - 27 = 6Ax - 7A + B \].

Step 6: Identify Coefficients

By comparing coefficients from both sides of the equation, we find two equations:1. \(6A = -24\)2. \(-7A + B = -27\).

Step 7: Solve for A

Solve the first equation for \(A\): \[ 6A = -24 \] Dividing both sides by 6 gives: \[ A = -4 \].

Step 8: Solve for B

Substitute \(A = -4\) into the second equation: \[-7(-4) + B = -27\] \[28 + B = -27\] Solving for \(B\), we get: \[B = -27 - 28 = -55\].

Step 9: Write the Final Partial Fraction Decomposition

Substitute the found values of \(A\) and \(B\) into the partial fractions: \[ \frac{-24x - 27}{(6x-7)^2} = \frac{-4}{6x-7} + \frac{-55}{(6x-7)^2} \].

Key Concepts

Rational Functions

A rational function is simply a fraction where the top (the numerator) and the bottom (the denominator) are both polynomials. It looks something like this: \[\frac{-24x - 27}{(6x - 7)^2}\]Here,

• The numerator is \(-24x - 27\), a simple polynomial.
• The denominator is \((6x - 7)^2\), which involves a polynomial raised to a power.

Rational functions are important because they appear frequently in calculus and algebra, especially when working with limits, series, or solving equations. To analyze them more smoothly, we often use techniques like partial fraction decomposition, which breaks down complex rational expressions into simpler fractions, making them easier to work with. This becomes very useful when integrating rational functions in calculus.

Repeating Linear Factors

When dealing with rational functions, you might encounter denominators with factors that repeat. In our example, the denominator is \((6x - 7)^2\). This means the linear factor \((6x - 7)\) appears twice. Repeating linear factors affect how we break down a rational expression into partial fractions. We represent each occurrence of the factor in the fraction separately, giving our partial fraction decomposition this form: \[\frac{A}{6x - 7} + \frac{B}{(6x - 7)^2}\]Here,

• The first term is associated with the single power of \((6x - 7)\).
• The second term deals with the repeated factor \((6x - 7)^2\).

By setting up the decomposition correctly, we ensure that our approach accounts for each occurrence of the repeating factor. This method is integral (pun intended!) when working towards simplifying or integrating rational expressions.

Coefficient Comparison

A core technique in solving partial fraction decompositions is comparing coefficients of like terms. Once you have the partial fraction setup like \[-24x - 27 = A(6x - 7) + B\]The next step is eliminating the denominator and expanding both sides if needed, as was done above. This gives us an expression involving the coefficients of \(x\) and constant terms that we can compare from both sides: \[-24x - 27 = 6Ax - 7A + B\]

• For the coefficients of \(x\), equate: \(6A = -24\)
• For the constant terms, equate: \(-7A + B = -27\)

Solving these equations gives the values of \(A\) and \(B\). This approach works because polynomials are equal if and only if their corresponding coefficients are equal. It's a straightforward way to solve for variables in algebra and ensure accuracy in partial fraction decomposition.