Problem 30 - Algebra and Trigonometry

Question

\(13-44=\) Find the partial fraction decomposition of the rational function. $$ \frac{x-4}{(2 x-5)^{2}} $$

Step-by-Step Solution

Answer

\( \frac{x-4}{(2x-5)^2} = \frac{1/2}{2x-5} - \frac{3/2}{(2x-5)^2} \).

Step 1: Identify the Problem Type

The problem asks for the partial fraction decomposition of the expression \( \frac{x-4}{(2x-5)^2} \). We need to express this as a sum of simpler fractions.

Step 2: Setup the Decomposition Structure

Given that the denominator \((2x-5)^2\) is a repeated linear factor, the decomposition will be of the form: \[ \frac{x-4}{(2x-5)^2} = \frac{A}{2x-5} + \frac{B}{(2x-5)^2}. \] Here, \(A\) and \(B\) are constants that we need to find.

Step 3: Clear the Denominator

Multiply both sides of the equation by \((2x-5)^2\) to get rid of the denominators: \[ x-4 = A(2x-5) + B. \] This equation must hold for all values of \(x\).

Step 4: Expand and Collect Like Terms

Expand \(A(2x-5)\): \[ A(2x-5) = 2Ax - 5A. \] So the equation becomes \[ x-4 = 2Ax - 5A + B. \]

Step 5: Compare Coefficients

To find \(A\) and \(B\), compare coefficients on both sides of the equation. For \(x\), we have \(1 = 2A\) (since there is \(x\) term) and for the constant term, we have \(-4 = -5A + B\).

Step 6: Solve for A and B

First, solve for \(A\) from the equation \(1 = 2A\) which implies \(A = \frac{1}{2}\). Substitute \(A = \frac{1}{2}\) into \(-4 = -5A + B\): \[ -4 = -5\left(\frac{1}{2}\right) + B \] which simplifies to \(-4 = -\frac{5}{2} + B\). Add \(\frac{5}{2}\) to both sides to find \(B\): \(-4 + \frac{5}{2} = B\), resulting in \(B = -\frac{8}{2} + \frac{5}{2} = -\frac{3}{2}\).

Step 7: Write the Final Decomposition

Substitute \(A\) and \(B\) back into the partial fraction decomposition: \[ \frac{x-4}{(2x-5)^2} = \frac{1/2}{2x-5} - \frac{3/2}{(2x-5)^2}. \]

Key Concepts

Rational Functions

Rational functions are fractions composed of polynomials, with one polynomial sitting in the numerator and another in the denominator. In our exercise, the rational function is \(\frac{x-4}{(2x-5)^2}\). These functions appear frequently in mathematics, science, and engineering because they can model many types of relationships and systems.

When decomposing rational functions into partial fractions, we want to break the expression down into simpler, more manageable pieces. This process helps us solve integrals more easily, among other applications. Understanding how to efficiently perform these decompositions is crucial in both calculus and algebra. To start, we consider the degree of the numerator versus the degree of the denominator. If the degree of the numerator is greater than or equal to that of the denominator, polynomial long division is needed. However, in our scenario, the numerator \(x-4\) is linear, and the denominator \((2x-5)^2\) is quadratic, making our setup straightforward for decomposition without division.

Repeated Linear Factors

In the world of rational functions, repeated linear factors are those which appear more than once in the denominator. Our function \(\frac{x-4}{(2x-5)^2}\) showcases this with the factor \((2x-5)^2\). Repeated linear factors influence the structure of the partial fraction decomposition significantly.

When faced with a repeated linear factor such as \((2x-5)^2\), each factor must have a corresponding partial fraction term in the decomposition. The general guideline is to include separate terms up to the power of the factor. Hence, for our repeated factor \((2x-5)^2\), the decomposition takes the form:

• \(\frac{A}{2x-5}\) for the first degree factor, and
• \(\frac{B}{(2x-5)^2}\) for the full repeated factor

The coefficients \(A\) and \(B\) must be determined so that the equation holds for all \(x\). Repeated linear factors can increase the complexity of the problem, but they offer a structured way to address these decompositions.

Coefficient Comparison

One of the key techniques for determining the constants \(A\) and \(B\) in a partial fraction decomposition is coefficient comparison. This method involves equating the numerator of the original rational function to the expanded form of the decomposed expression.

After clearing the denominators, our equation becomes \(x-4 = A(2x-5) + B\). By expanding and organizing like terms, we align the expression with the original equation:

• For the \(x\) terms, we equate the coefficients: \(1 = 2A\), solving gives \(A = \frac{1}{2}\).
• For the constant terms, we have \(-4 = -5A + B\), and substitute \(A\) to solve for \(B\).

The coefficient comparison method is highly effective as it provides a systematic path to identifying the values of constants, ensuring the partial fraction accurately reconstructs the original rational function.