Question

Set up the appropriate form of the partial fraction decomposition for the following expressions. Do not find the values of the unknown constants. $$\frac{x}{\left(x^{4}-16\right)^{2}}$$

Short answer

Answer: The expression \(\frac{x}{(x^4-16)^2}\) should be rewritten in the form of partial fraction decomposition as: \(\frac{x}{(x^4-16)^2} = \frac{A}{x-2} + \frac{B}{(x-2)^2} + \frac{C}{x+2} + \frac{D}{(x+2)^2} + \frac{Ex+F}{x^2+4}\).

Step-by-step solution

Factor the Denominator

First, we need to factor the denominator. We can factor the expression \(x^4 - 16\) as a difference of squares: $$x^4 - 16 = (x^2 - 4)(x^2 + 4)$$ Furthermore, we can factor \((x^2 - 4)\) as another difference of squares: $$(x^2 - 4) = (x-2)(x+2)$$ So our denominator becomes: $$(x^4 - 16)^2 = [(x-2)(x+2)(x^2+4)]^2$$

Set Up Partial Fraction Decomposition

Now that we have factored the denominator, we can set up the appropriate form of partial fraction decomposition. Since the factors \((x-2)\), \((x+2)\), and \((x^2+4)\) are all raised to the power of 2, we will have two fractions for each linear factor and one fraction for the quadratic factor: $$\frac{x}{(x^4-16)^2} = \frac{A}{x-2} + \frac{B}{(x-2)^2} + \frac{C}{x+2} + \frac{D}{(x+2)^2} + \frac{Ex+F}{x^2+4}$$ In this decomposition, the unknown constants are \(A, B, C, D, E,\) and \(F\).

Key concepts

Factoring Denominators

Factoring denominators is a fundamental step in simplifying complex algebraic expressions, especially when performing partial fraction decomposition. This process involves breaking down a polynomial into a product of simpler polynomials that are easier to work with. For instance, in the exercise, the denominator is a polynomial given by the expression \(x^4 - 16\). The first step is identifying this as a difference of squares, a common pattern where two perfect squares are subtracted from each other.

By recognizing \(x^4\) as \( (x^2)^2 \) and \(16\) as \(4^2\), the expression simplifies to \( (x^2 - 4)(x^2 + 4) \), following the formula \( a^2 - b^2 = (a - b)(a + b) \). Further breaking down the \( x^2 - 4 \) term, we get the linear factors \( (x - 2)(x + 2) \).

Factoring denominators is essential because it allows us to convert a rational function into a sum of simpler fractions, making it more tractable for integration, differentiation, or finding limits. Without this crucial step, it becomes significantly more challenging to solve complex algebraic expressions.

Difference of Squares

The difference of squares is a mathematical pattern used to break down expressions that subtract one square number from another square number. The general formula is \( a^2 - b^2 = (a + b)(a - b) \). In the context of partial fraction decomposition, recognizing a difference of squares allows us to factor the given expression further.

For example, in our problem, the expression \(x^4 - 16\) is initially factored into \(x^2 - 4\) and \(x^2 + 4\). We can then further factor \( x^2 - 4 \) because it also represents a difference of squares (where \(a = x\) and \(b = 2\)). The result is the product of linear factors \( (x - 2)(x + 2) \).

Understanding the difference of squares is imperative for simplifying algebraic expressions and is particularly useful in calculus and higher-level mathematics when dealing with more complex integrals and related algebraic manipulations.

Linear Factors

Linear factors are the building blocks of polynomials and represent expressions of the first degree, which means they have the general form \( ax + b \) where \( a \) and \( b \) are constants, and \(a \eq 0\). When we talk about partial fraction decomposition, after factoring a polynomial, we seek to break up our expression into fractions that have these linear factors in the denominator.

In our problem, after factoring the difference of squares, we identified two linear factors: \( (x-2) \) and \( (x+2) \). Each distinct linear factor in the denominator of the original expression will correspond to a separate term in the partial fraction decomposition, often resulting in simpler, more manageable components for further mathematical operations.

Quadratic Factors

Quadratic factors are expressions of the second degree, in the form \( ax^2 + bx + c \) where \( a \) is a non-zero constant. While linear factors are essential for understanding the basics of factoring, quadratic factors play a crucial role when the polynomial cannot be factored into purely linear components.

In the exercise, after factoring out the linear terms, we are left with a quadratic term \( x^2 + 4 \) that cannot be further factored over the real numbers since it does not have real roots. This quadratic factor will yield a unique term in the partial fraction decomposition, typically involving both x and a constant in the numerator, as in the \( (Ex + F) \)/(x^2 + 4) term of our decomposition formula. Recognizing and dealing with quadratic factors is crucial for accurate and complete partial fraction decomposition.

Unknown Constants

Unknown constants in partial fraction decomposition are placeholders representing the coefficients we need to determine to fully specify the decomposed fractions. They are essential because they allow us to express a complex fraction as a sum of simpler fractions, each of which can be solved for or integrated more easily.

In the given exercise, unknown constants are denoted as \( A, B, C, D, E, \) and \( F \). After setting up the equations based on the given expression, one would typically solve for these constants by equating coefficients or using other methods such as substitution. The exercise specifically avoids finding the values of these constants but understanding their role is critical. Once the unknown constants are determined, they provide the full decomposition of the original expression, which can then be used for various purposes like simplification, integration, or solving equations.