Problem 12 - 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. $$\frac{1}{\left(x^{3}-1\right)\left(x^{2}-1\right)}$$

Step-by-Step Solution

Answer

Partial fraction form: \( \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x+1} + \frac{Dx+E}{x^2+x+1} \).

Step 1: Identify the Factors in the Denominator

The given function has the denominator \( (x^3 - 1)(x^2 - 1) \). We need to factor each of these terms.\( x^3 - 1 \) can be factored as \((x - 1)(x^2 + x + 1)\) using the difference of cubes formula. \( x^2 - 1 \) factors into \((x - 1)(x + 1)\) as the difference of squares.

Step 2: Write the Factored Form of the Denominator

Using the factored forms, the denominator becomes \((x - 1)^2(x + 1)(x^2 + x + 1)\).

Step 3: Setup the Partial Fraction Decomposition

For each factor, write corresponding terms for the partial fraction decomposition: 1. For \((x-1)^2\), include terms \(\frac{A}{x-1} + \frac{B}{(x-1)^2}\).2. For \(x+1\), include a term \(\frac{C}{x+1}\).3. For \(x^2+x+1\), include a term \(\frac{Dx+E}{x^2+x+1}\) since it is an irreducible quadratic factor.

Step 4: Final Formulate the Partial Fraction Decomposition

The partial fraction decomposition is structured as: \[ \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x+1} + \frac{Dx+E}{x^2+x+1} \] where \( A, B, C, D, \) and \( E \) are constants to be determined.

Key Concepts

Polynomial Factoring

Polynomial factoring is like finding the building blocks of a polynomial expression. By breaking it down into simpler expressions or factors, you gain insight into how the polynomial behaves. A factor is a smaller expression that multiplies with others to form the original polynomial. This process is essential in simplifying expressions and solving equations. For instance, to factor the polynomial \( x^3 - 1 \), recognize it as a difference of cubes. Use the formula for difference of cubes: \( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \). Here, set \( a = x \) and \( b = 1 \), resulting in \( (x-1)(x^2+x+1) \).
Similarly, for \( x^2-1 \), apply the difference of squares factoring: \( a^2-b^2 = (a-b)(a+b) \). Setting \( a = x \) and \( b = 1 \), it factors to \( (x-1)(x+1) \). Understanding these patterns allows us to handle more complex polynomial equations.

Difference of Cubes

The difference of cubes involves two numbers or expressions cubed, subtracted from each other, simplified with a specific formula. For any two numbers \( a \) and \( b \), the expression \( a^3 - b^3 \) factors into \( (a-b)(a^2 + ab + b^2) \).
This factoring is fundamental, especially when simplifying the denominator of partial fractions. In our exercise, the expression \( x^3 - 1 \) turns into \( (x-1)(x^2+x+1) \), which helps set up partial fraction decomposition. Notice how the simple linear factor \( (x - 1) \) appears alongside the quadratic expression \( (x^2 + x + 1) \). Recognizing and utilizing the difference of cubes enables handling complex algebraic problems more efficiently.

Difference of Squares

The difference of squares formula is a crucial tool in polynomial factoring. It applies to expressions of the form \( a^2 - b^2 \), which factor into \( (a-b)(a+b) \). This type of factoring is straightforward but very powerful in breaking down polynomials.
In our example, the expression \( x^2 - 1 \) can be identified as a difference of squares. Applying the formula directly, it breaks down to \( (x-1)(x+1) \). This decomposition illuminates the inner structure of the polynomial, assisting in further calculations like partial fraction decomposition. Such simplifications are invaluable in solving polynomial equations and performing calculus operations.

Partial Fractions

Partial fractions involve expressing a complex fraction as a sum of simpler rational expressions. This process simplifies the integration and differentiation of rational functions. In our exercise, the goal is to decompose the function \( \frac{1}{(x^3-1)(x^2-1)} \) into simpler fractions.
After factoring the denominator, \( (x - 1)^2(x + 1)(x^2 + x + 1) \), write the partial fraction decomposition. Each factor corresponds to a term in the decomposition. For the linear term \((x-1)^2\), include two terms: \( \frac{A}{x-1} \) and \( \frac{B}{(x-1)^2} \). For \( x+1 \), add \( \frac{C}{x+1} \). Finally, for the irreducible quadratic \( x^2+x+1 \), introduce \( \frac{Dx+E}{x^2+x+1} \).
This setup makes complex rational functions manageable, particularly when solving integrals. It breaks down complicated expressions into simpler forms, paving the way for easier computations.