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

Decompose as \(\frac{A}{x-1} + \frac{B}{x+1} + \frac{Cx+D}{x^2+1}\).

Step 1: Factor the Denominator

The first step in partial fraction decomposition is to factor the denominator \(x^4 - 1\). We recognize this as a difference of squares: \(x^4 - 1 = (x^2 - 1)(x^2 + 1)\). Further factor \(x^2 - 1\) to get \((x - 1)(x + 1)\). Thus, the full factorization is \((x - 1)(x + 1)(x^2 + 1)\).

Step 2: Set Up the Partial Fraction Decomposition

Since the factors \((x - 1)\), \((x + 1)\) are linear, and \((x^2 + 1)\) is quadratic with no real roots, we'll set up the decomposition as follows:\[\frac{1}{(x - 1)(x + 1)(x^2 + 1)} = \frac{A}{x - 1} + \frac{B}{x + 1} + \frac{Cx + D}{x^2 + 1}\]Here, \(A\), \(B\), \(C\), and \(D\) are coefficients to be determined.

Key Concepts

Factoring Polynomials

Factoring polynomials is a fundamental concept in algebra that involves breaking down a polynomial into a product of simpler polynomials. This technique is crucial in various mathematical processes, such as solving equations and simplifying expressions. The process begins by identifying common factors among the terms or recognizing patterns, such as

• Difference of squares
• Perfect square trinomials
• Sum or difference of cubes

For instance, when we see the polynomial \(x^4 - 1\), a common technique involves recognizing it as a difference of squares. By understanding and applying these techniques, factoring polynomials becomes a manageable task, paving the way for processes like partial fraction decomposition.

Difference of Squares

The difference of squares is a special factoring formula used when a polynomial is in the form \(a^2 - b^2\). This formula can be represented as \(a^2 - b^2 = (a - b)(a + b)\). It is called 'difference' because the terms are only subtracted, and 'squares' because both terms are perfect squares. In this context:

• The term \(x^4 - 1\) fits the pattern because \(x^4\) is \((x^2)^2\) and 1 is a square.
• Thus, we can write \(x^4 - 1 = (x^2)^2 - 1^2\).
• Applying the difference of squares formula gives us \((x^2 - 1)(x^2 + 1)\).

This key step not only simplifies the expression but also sets up further opportunities for decomposing into partial fractions.

Denominator Factorization

Denominator factorization is essential in the process of partial fraction decomposition. It involves breaking down the denominator of a rational function into its simplest factors. This step is critical for setting up the decomposition formula.In the example of \(\frac{1}{x^4 - 1}\), the factorization begins by recognizing \(x^4 - 1\) as a difference of squares, which results in a further factorization into \((x^2 - 1)(x^2 + 1)\).From there, the factor \(x^2 - 1\) is recognized as another difference of squares, giving us \((x - 1)(x + 1)\).Thus, the complete factorization of the denominator is \((x - 1)(x + 1)(x^2 + 1)\). This orderly breakdown is crucial for the next step in forming the partial fractions.

Linear and Quadratic Factors

In the context of partial fraction decomposition, understanding the nature of the factors is essential. Factors can be linear or quadratic:Linear factors are those in the form of \(ax + b\). They are straightforward to deal with in decomposition. For example, \(x-1\) and \(x+1\) are linear factors of the polynomial \(x^4 - 1\).Quadratic factors are in the form of \(ax^2 + bx + c\), which may not have real roots. The factor \(x^2 + 1\) is a quadratic factor.In partial fraction decomposition:

• Each linear factor gets a simple constant numerator. So, \(\frac{A}{x-1}\) and \(\frac{B}{x+1}\) represent fractions with linear denominators.
• For irreducible quadratic factors, the numerator is generally linear. Thus, \(\frac{Cx + D}{x^2 + 1}\) is used, where \(Cx + D\) accounts for possible variation within that portion of the decomposition.

Knowing whether a factor is linear or quadratic helps in setting up the correct form for the partial fraction, ensuring that the coefficients can later be determined accurately.