Problem 19 - College Algebra

Question

Write the partial fraction decomposition of each rational expression. $$\frac{4 x^{2}-7 x-3}{x^{3}-x}$$

Step-by-Step Solution

Answer

The partial fraction decomposition of the given rational expression \(\frac{4 x^{2}-7 x-3}{x^{3}-x}\) can be represented as \(\frac{A}{x} + \frac{B}{x-1} + \frac{C}{x+1}\), where A, B and C will be the values determined during the solution steps.

Step 1: Simplify the equation

Given the equation is \(\frac{4 x^{2}-7 x-3}{x^{3}-x}\), the denominator can be factored into \(x(x^2 - 1)\). Now, the equation looks like: \(\frac{4x^2 -7x -3}{x(x^2 - 1)}\).

Step 2: Factorize the denominator further

The term \(x^2 - 1\) is a difference of squares and can be factorized as \((x-1)(x+1)\). The equation can now be represented as \(\frac{4x^2 -7x -3}{x(x-1)(x+1)}\).

Step 3: Express the equation with undetermined coefficients

The equation is broken down into simpler fractions as: \(\frac{A}{x} + \frac{B}{x-1} + \frac{C}{x+1}\) where A, B and C are coefficients that we need to determine.

Step 4: Equal the original and the decomposed fractions

By equating the given fraction and the decomposed fraction, we get \(4x^2 -7x -3 = A(x^2 -1) + Bx(x+1) + Cx(x-1)\), where \(A\), \(B\) and \(C\) are constants.

Step 5: Solve for the coefficients A, B and C

By taking suitable values of \(x\) which makes the terms 0, it's possible to solve for the coefficients \(A\), \(B\), and \(C\). Take \(x = 0, 1, -1\), you can solve the resulting linear system of equations to find the values of \(A\), \(B\), and \(C\).

Step 6: Substitute the values of A, B and C back into the fractions

Replace \(A\), \(B\), and \(C\) in \(\frac{A}{x} + \frac{B}{x-1} + \frac{C}{x+1}\) with their respective values to obtain the final partial fraction decomposition.

Key Concepts

Rational Expressions

Rational expressions are like fractions, but they involve polynomials in both the numerator and the denominator. To understand these expressions better, think of them as division problems between two polynomials. This means that a rational expression must be defined in the same way fractions are, meaning the denominator cannot equal zero. In algebra, these expressions are incredibly important as they represent a wide variety of scenarios.

• Just like fractions, we can add, subtract, multiply, and divide rational expressions.
• We need to be cautious to ensure the denominator is never zero, because division by zero is undefined.

Understanding and manipulating rational expressions provide the foundation for working with more complex algebraic structures. And it plays a pivotal role when it comes to partial fraction decomposition, where one complex rational expression is expressed as a sum of simpler fractions.

Factoring Polynomials

Factoring polynomials is a critical step when dealing with rational expressions, especially for simplifying them. Essentially, factoring is the process of breaking down a polynomial into a product of simpler polynomials that, when multiplied together, give back the original polynomial.
To factor a polynomial:

• Look for common factors in terms and simplify first.
• Use techniques like factoring by grouping or applying formulas like the difference of squares.
• In some cases, recognizing patterns or using the quadratic formula may be necessary.

In the context of our exercise, factoring was used to break down the denominator from \(x^3 - x\) into \(x(x-1)(x+1)\). This step simplifies working with the rational expression and sets up the equation for partial fraction decomposition. Factoring is fundamental in simplifying complex expressions and finding solutions more efficiently.

Algebraic Fractions

Algebraic fractions, simply put, are fractions where the numerator and/or the denominator contain algebraic expressions. These are important when performing operations such as addition or subtraction, and they become particularly useful once factored properly.
When dealing with algebraic fractions:

• Ensure the polynomial in the denominator doesn’t equal zero to avoid undefined values.
• Common denominators are essential for adding or subtracting these fractions.
• Always simplify the fractions, if possible, to make calculations easier.

In our problem, after factoring, each component of the denominator (\(x\), \(x-1\), \(x+1\)) was turned into a separate fraction with unknown coefficients. This step is pivotal in converting a complex expression into manageable parts for partial fraction decomposition.

Solving Linear Equations

Solving linear equations is a key concept, especially when determining unknown constants in partial fraction decomposition. A linear equation involves variables raised to the first power, making them straightforward to solve.
Steps to solve such equations:

• Isolate the variable on one side of the equation. Doing so might require operations like addition, subtraction, multiplication, or division.
• Substitute values to check if you’ve found the correct solution.
• Use strategic values for variables to simplify calculations, often aligning these values with steps that eliminate terms to solve for specific variables.

In the exercise, setting \(x = 0, 1, -1\) allowed us to create simple linear equations that could be solved for the constants \(A\), \(B\), and \(C\). Solving these equations accurately ensures the correctness of the partial fraction decomposition.