Problem 10 - Algebra and Trigonometry

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

The partial fraction decomposition is \( \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x+1} + \frac{Dx + E}{x^2+x+1} \).

Step 1: Factor the Polynomial Denominator

First, we need to factor each polynomial in the denominator of the expression \(\frac{1}{(x^3-1)(x^2-1)}\). The expression \(x^3-1\) can be factored using the difference of cubes formula: \(x^3-1 = (x-1)(x^2+x+1)\).The expression \(x^2-1\) is a difference of squares: \(x^2-1 = (x-1)(x+1)\). Thus, the denominator becomes \((x-1)^2(x+1)(x^2+x+1)\).

Step 2: Set Up the Partial Fraction Decomposition

Next, we set up the partial fraction decomposition based on the factors. Since \((x-1)^2\) is a repeated linear factor, using different notations, we allocate separate terms for each power:\[ \frac{A}{x-1} + \frac{B}{(x-1)^2} \]For the linear factor \((x+1)\), we add:\[ \frac{C}{x+1} \]For the irreducible quadratic factor \((x^2+x+1)\), we add:\[ \frac{Dx + E}{x^2+x+1} \]The full partial fraction decomposition is:\[ \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x+1} + \frac{Dx + E}{x^2+x+1} \]

Key Concepts

Polynomial Factorization

Polynomial factorization is a method of expressing a polynomial as the product of its factors. Imagine splitting it into simpler polynomials whose values multiply together to give the original polynomial. In algebra, this is a crucial step because it helps simplify expressions and solve equations.

A polynomial like \(x^3 - 1\) can often be broken down into factors by using known formulas or techniques. This helps to identify each individual component or factor of the polynomial that is a part of a multiplied expression.

• It simplifies working with complex algebraic fractions or partial fractions.
• Allows solving polynomial equations by setting each factor to zero.
• Leads to easier integrations and derivations in calculus problems.

Mastering polynomial factorization is key to navigating through other mathematical concepts seamlessly.

Difference of Squares

The difference of squares is a special algebraic identity used to factor expressions of the form \(a^2 - b^2\). This can be rewritten as \((a-b)(a+b)\). It represents a shortcut for factorization that avoids expanding the whole expression.

For instance, in the exercise, \(x^2 - 1\) reduces to the factors \((x-1)(x+1)\). Recognizing such patterns quickly is a huge advantage in math:

• Makes simplification straightforward by using the identities.
• Very useful when working on equations and inequalities.
• Helps in understanding properties of algebraic structures.

Understanding the difference of squares not only aids in factorization but also builds a foundation for learning other algebraic identities.

Difference of Cubes

The Difference of Cubes is another special algebraic identity for expressions of the form \(a^3 - b^3\). This can be broken down as \((a-b)(a^2 + ab + b^2)\). This technique is essential for simplifying cubic expressions and helps in various problem-solving scenarios.

Applying this to \(x^3 - 1\), it becomes \((x-1)(x^2+x+1)\). Using differences of cubes allows tackling cubic equations efficiently:

• Facilitates easier transformations into simpler polynomials.
• Essential in derivations needed for calculus problems.
• Useful for exploring number theory and polynomial properties.

Mastering this identity serves as a building block for dealing with complex algebraic expressions and enhances understanding of polynomial behavior.

Irreducible Quadratic Factor

An irreducible quadratic factor is a quadratic expression that cannot be factored further using real numbers. This means it cannot be expressed as the product of two linear factors having real coefficients.

In our case, the factor \(x^2+x+1\) is such an irreducible factor. In these cases, it often requires setting up separate linear terms when doing partial fraction decomposition.

• These factors indicate no real roots exist; they stay in quadratic form.
• They are sometimes involved in complex roots in the complex number plane.
• In partial fraction decomposition, they require terms of \(Ax+B\) form over them.

Understanding irreducible quadratic factors is crucial when handling partial fractions, ensuring that expressions are properly broken down in the solution process.