Problem 23 - Precalculus Mathematics for Calculus

Question

Find the partial fraction decomposition of the rational function. $$\frac{8 x-3}{2 x^{2}-x}$$

Step-by-Step Solution

Answer

\(\frac{3}{x} + \frac{2}{2x - 1}\)

Step 1: Factor the Denominator

First, factor the denominator of the function \(2x^2 - x\). Notice that we can factor out an \(x\), giving the expression \(x(2x - 1)\).

Step 2: Setup the Partial Fraction Decomposition

With the factored denominator \(x(2x - 1)\), express \(\frac{8x - 3}{2x^2 - x}\) as the sum of two fractions: \(\frac{A}{x} + \frac{B}{2x - 1}\), where \(A\) and \(B\) are constants to be determined.

Step 3: Clear the Denominator

Multiply through by the common denominator \(x(2x - 1)\) to eliminate the fractions, leading to the equation:\[8x - 3 = A(2x - 1) + Bx\]

Step 4: Expand and Collect Like Terms

Expand the right side of the equation:\[8x - 3 = 2Ax - A + Bx\]Combine like terms:\[8x - 3 = (2A + B)x - A\]

Step 5: Equate and Solve for Coefficients

By comparing coefficients, we have the system of equations:\[2A + B = 8 \-A = -3\]From \(-A = -3\), solve for \(A\):\[A = 3\]Substitute \(A = 3\) into the first equation:\[2(3) + B = 8\]\[6 + B = 8\]\[B = 2\]

Step 6: Write the Final Decomposition

Substitute \(A = 3\) and \(B = 2\) into the partial fractions:\[\frac{8x - 3}{2x^2 - x} = \frac{3}{x} + \frac{2}{2x - 1}\]

Key Concepts

Rational Function

A rational function is essentially a fraction where both the numerator and the denominator are polynomials. For example, in the function \(\frac{8x - 3}{2x^2 - x}\), the polynomial \(8x - 3\) is the numerator and \(2x^2 - x\) is the denominator. The key characteristic of a rational function is that it behaves similarly to a fraction of integers. It can be broken down into simpler parts, which is especially useful when working with calculus operations, like integration.
This breakdown involves expressing the rational function as a sum of simpler fractions, often called partial fractions. This approach allows for easier manipulation and calculation of the integral or other operations on the function. Each partial fraction is simpler because either its denominator is a linear factor or a higher-degree polynomial that cannot be further factored over the real numbers.
Understanding rational functions and their structure can help to simplify complex problems by reducing them to more manageable pieces.

Factoring Polynomials

Factoring polynomials involves breaking down a complex polynomial into a product of simpler polynomials. In the partial fraction decomposition process, the first step is to factor the polynomial in the denominator. For our exercise, the polynomial in the denominator is \(2x^2 - x\). By taking out the greatest common factor (GCF), which is \(x\), we simplify it to \(x(2x - 1)\).
Factoring is essential because it allows us to set up the partial fractions. The denominators of these partial fractions will be these factored expressions, which are easier to manage. It's crucial to remember:

• Always look for the GCF first before attempting other factoring methods.
• Factoring sets the stage for converting a rational expression into a sum of simpler fractions.

This step makes the problem more approachable and forms the foundation for the following steps in partial fraction decomposition.

Comparing Coefficients

Comparing coefficients is a method used to determine the values of unknown constants in an equation involving polynomials. Once we have cleared the denominators and expanded the polynomial equation, we equate coefficients of corresponding powers of \(x\) from both sides of the equation.
In the context of partial fraction decomposition, let's consider the expanded equation: \(8x - 3 = (2A + B)x - A\). Here, the terms involving \(x\) on both sides are \(8x\) and \((2A + B)x\). By matching these coefficients of \(x\), we form the equation \(2A + B = 8\). Similarly, for the constant terms, \(-3\) and \(-A\), we get \(-A = -3\).
This systematic approach to solving for unknowns by comparing coefficients is crucial:

• It provides a straightforward pathway to finding the constants \(A\) and \(B\).
• Each pair of like terms gives us a separate equation.

By solving these resulting simple equations, we find the required constants to finalize the partial fraction decomposition.