Problem 7 - 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. \(\frac{x^{3}-4 x^{2}+2}{\left(x^{2}+1\right)\left(x^{2}+2\right)}\)

Step-by-Step Solution

Answer

The decomposition is \(\frac{Ax+B}{x^2+1} + \frac{Cx+D}{x^2+2}\).

Step 1: Identify the Denominator

Identify the factors of the denominator \((x^2+1)(x^2+2)\). Since both are irreducible quadratic factors, we will consider them separately in the decomposition.

Step 2: Set Up Partial Fraction Form

Express the rational function as a sum of fractions, each with the irreducible quadratic in the denominator as follows: \(\frac{x^3 - 4x^2 + 2}{(x^2+1)(x^2+2)} = \frac{Ax+B}{x^2+1} + \frac{Cx+D}{x^2+2}\).

Step 3: Explain the Coefficient Setup

Each irreducible quadratic factor in the denominator \((x^2+1)\) and \((x^2+2)\) will have a corresponding numerator that is a linear polynomial \((Ax+B)\) and \((Cx+D)\), respectively. This setup accounts for all possible degrees of numerators needed to match the degree of the original polynomial in its expanded form without repeating linear factors.

Key Concepts

Irreducible Quadratic Factors

In the world of partial fraction decomposition, irreducible quadratic factors are quite important. An irreducible quadratic factor is a quadratic polynomial that cannot be factored further over the real numbers or does not have real roots. In simpler words, you can't break these quadratic expressions into simpler linear factors if you stay away from complex numbers. For the problem at hand, the denominator is made up of two such quadratics: \(x^2+1\) and \(x^2+2\). Both of these expressions are irreducible over the real numbers because neither has real solutions. Each factor represents a term in the partial fraction decomposition, and we treat them separately. Understanding that these can't be reduced helps us understand why each must have a linear polynomial numerator.

Linear Polynomial Numerators

Once we identify the irreducible quadratic factors in the denominator, the next task is to decide on the form of the numerators. In partial fraction decomposition, a linear polynomial numerator is often used in conjunction with an irreducible quadratic denominator. This means if you have an irreducible quadratic factor like \(x^2 + 1\), its corresponding numerator should be something simple yet flexible enough to capture variations. Hence, it takes the form \(Ax + B\). This approach ensures that each component has the potential to balance out the terms in the rational function. Each linear polynomial can adjust using constants \(A\) and \(B\). They act like dials, which later get determined by comparing coefficients or solving equations, ensuring that the original rational expression is faithfully represented.

Rational Function

A rational function is a fraction where both the numerator and denominator are polynomials. In our exercise, the rational function begins with \(\frac{x^3 - 4x^2 + 2}{(x^2+1)(x^2+2)}\). Rational functions can often be complicated, but breaking them down into simpler parts can help interpret their behavior and solve equations involving them. Partial fraction decomposition is a technique applied to rational functions to express them as simpler, easier-to-invert components. This process is particularly useful for integration and solving differential equations. Understanding the structure of a given rational function is the first step in decomposing it into partial fractions. The procedure simplifies the function, separating complex interactions into individual components.

Polynomial Division

Polynomial division comes into play when dealing with rational functions. Although not directly needed in all partial fraction decompositions, it ensures a proper rational expression before decomposition starts. A proper rational function has a numerator with a lesser degree than the denominator. If our original rational function had a numerator with a higher degree than the denominator, then polynomial division would let us rewrite it in a simpler form. By dividing the numerator by the denominator, this method provides a quotient plus a remainder, ensuring the rational function is decomposable. In our specific problem, the numerator's degree (3) is less than the combined degree of the denominator (4), so polynomial division is unnecessary. Nevertheless, understanding polynomial division helps in cases where further simplification is needed before employing partial fractions.