Problem 8 - Precalculus Mathematics for Calculus

Question

Write the form of the partial fraction decomposition of the function (as in Example 4 ). Do not determine the numerical values of the coefficients. $$\frac{1}{x^{4}-1}$$

Step-by-Step Solution

Answer

\(\frac{1}{x^4-1} = \frac{A}{x-1} + \frac{B}{x+1} + \frac{Cx+D}{x^2+1}\)

Step 1: Factor the Denominator

First, observe the expression in the denominator, \(x^4 - 1\). Recognize this can be factored using the difference of squares. Begin by factoring it as \((x^2 - 1)(x^2 + 1)\). Then, continue by factoring \(x^2 - 1\) further into \((x-1)(x+1)\). The fully factored form is \((x-1)(x+1)(x^2+1)\).

Step 2: Set Up the Partial Fractions

Now that the denominator is factored, set up the partial fraction decomposition. For each factor of the denominator, write a corresponding fraction. The linear factors \((x-1)\) and \((x+1)\) contribute terms with constant numerators \(A\) and \(B\). The irreducible quadratic \((x^2+1)\) contributes a term with a linear numerator \(Cx + D\). The setup is: \(\frac{1}{x^4-1} = \frac{A}{x-1} + \frac{B}{x+1} + \frac{Cx+D}{x^2+1}\).

Key Concepts

Factoring Polynomials

Before diving into partial fraction decomposition, it's essential to understand how to factor polynomials. Factoring is breaking down a complex expression into simpler pieces, usually products of polynomials of lower degrees. When you encounter a polynomial like \(x^4 - 1\), the goal is to express it as a product of smaller polynomials. This makes operations like integration and partial fraction decomposition much more manageable.

To factor \(x^4 - 1\), we start by recognizing it as a "difference of two powers." We notice this can be rewritten as \( (x^2)^2 - 1^2 \). By applying the difference of squares formula \(a^2 - b^2 = (a-b)(a+b)\), we express it as \((x^2 - 1)(x^2 + 1)\).

Next, we focus on \(x^2 - 1\), which is another difference of squares because \(x^2 - 1^2\) can be factored to \((x-1)(x+1)\). After completing these steps, \(x^4 - 1\) is fully factored as \((x-1)(x+1)(x^2+1)\).

Difference of Squares

The difference of squares is a critical concept in algebra used to factor expressions of the form \(a^2 - b^2\). It is one of the most commonly used factoring techniques and applies when an expression can be written as a subtraction between two perfect squares.

The formula is simple: \(a^2 - b^2 = (a-b)(a+b)\). This shows how the square difference can be split into two linear factors.

Applying this method transforms complex polynomials into simpler components. For instance, if we apply it to \(x^4 - 1\), we set \(a = x^2\) and \(b = 1\) to get \((x^2 - 1)(x^2 + 1)\). This technique reduces many complex expressions to simpler forms, making them easier to manipulate in calculus and algebraic operations.

Irreducible Quadratic

When you have an expression like \(x^2 + 1\), it's essential to consider whether it can be factored further. In the context of real numbers, \(x^2 + 1\) is what we refer to as an "irreducible quadratic." This means it cannot be factored into two linear factors with real coefficients.

The term irreducible quadratic is important in partial fraction decomposition. Unlike the difference of squares, where further factoring is possible, an irreducible quadratic must be approached differently. It remains as-is because it's not possible to find real-valued roots for \(x^2 + 1\).

In partial fraction decomposition, an irreducible quadratic in the denominator requires a linear expression in the numerator, like \(Cx + D\). This ensures that all potential polynomial divisions are considered, making it an essential part of breaking down complex rational expressions into simpler, more integrable parts.