Problem 22 - 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

The partial fraction decomposition is \(\frac{3}{x} + \frac{2}{2x-1}\).

Step 1: Factor the Denominator

The first step to decomposing a rational function into partial fractions is to factor the denominator. Given the function \(\frac{8x-3}{2x^2-x}\), we can factor the denominator as follows:\[2x^2 - x = x(2x-1)\]This shows that the denominator factors into distinct linear factors \(x\) and \(2x-1\).

Step 2: Set Up Partial Fraction Decomposition

Now that we have the factors of the denominator, we set up the form for the partial fraction decomposition. For each distinct linear factor, we use a term like \(\frac{A}{x}\) and \(\frac{B}{2x-1}\). Thus, the expression becomes:\[\frac{8x-3}{2x^2 - x} = \frac{A}{x} + \frac{B}{2x-1}\]

Step 3: Clear the Denominator

Multiply through by the common denominator \(2x^2-x\) to clear the fractions and obtain an equation involving polynomials:\[8x - 3 = A(2x - 1) + Bx\]

Step 4: Expand and Simplify

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

Step 5: Match Coefficients

To find the values of \(A\) and \(B\), match the coefficients of the terms on both sides:1. For the \(x\) term: \(2A + B = 8\)2. For the constant term: \(-A = -3\)From the second equation, \(A = 3\). Substitute into the first equation:\(2(3) + B = 8\)\(6 + B = 8\)\(B = 2\).

Step 6: Write the Final Decomposition

Now that we have determined \(A = 3\) and \(B = 2\), substitute these values back into the partial fraction form:\[\frac{8x-3}{2x^2 - x} = \frac{3}{x} + \frac{2}{2x-1}\]

Key Concepts

Rational Functions

Rational functions are fractions where both the numerator and the denominator are polynomials. They often look like \( \frac{P(x)}{Q(x)} \), where both \(P(x)\) and \(Q(x)\) are polynomials. These types of functions can be tricky to deal with, but breaking them down into simpler parts through partial fraction decomposition can be very helpful. This process simplifies the function into a sum of simpler fractions, making integration or other operations easier.

• The numerator \(P(x)\) is the polynomial on top.
• The denominator \(Q(x)\) is the polynomial at the bottom.
• The partial fraction decomposition is only applicable when the degree of the numerator is less than the degree of the denominator.

Understanding how to decompose rational functions into partial fractions can greatly simplify solving equations and integrating complex expressions.

Polynomials

Polynomials are expressions comprised of variables and coefficients, structured from terms or monomials. They are expressed in the form of \(a_nx^n + a_{n-1}x^{n-1} + ... + a_1x + a_0\), where the coefficients \(a_0, a_1, ..., a_n\) are constants.

• Each term in a polynomial has a variable raised to a power.
• The highest power of the variable in the polynomial is called the degree of the polynomial.

In the case of rational functions, understanding polynomials is crucial because both the numerator and the denominator are polynomials. Such knowledge allows for tasks like factoring and matching coefficients, which are essential steps in solving problems involving rational functions.

Linear Factors

A linear factor is a polynomial of degree one, which takes the form \(ax + b\). Linear factors are the simplest type of factors, and often appear as the result of factoring polynomials.Transforming the denominator into a product of linear factors simplifies the process of partial fraction decomposition. For example, in the exercise given, the factoring of \(2x^2 - x\) results in linear factors \((x)\) and \((2x-1)\). With these factors, we can decompose the rational function into a series of simpler fractions.

• Linear factors help in breaking down complex fractions.
• They pave the way for easily finding constants in partial fraction decomposition.

Using linear factors makes solving equations and performing operations on fractions more manageable.

Factoring

Factoring is a mathematical process used to break down a polynomial into simpler components, or 'factors,' that can be multiplied to obtain the original polynomial. It is an essential step in partial fraction decomposition, as it allows us to express a polynomial as a product of linear factors.By factoring the polynomial, we can determine the roots or intercepts, and use them to rewrite the denominator into a series of simpler linear factors. For instance, in the problem provided, the denominator \(2x^2 - x\) was factored into \((x)(2x-1)\).

• Factoring is crucial for simplifying rational functions into partial fractions.
• It prepares the rational function for simpler manipulation and solution finding.

In essence, mastering the factoring process enhances your ability to solve rational equations efficiently and enables easier calculation of integrals or differences.