Problem 22 - Algebra and Trigonometry

Question

write the partial fraction decomposition of each rational expression. $$\frac{x}{(x+1)^{2}}$$

Step-by-Step Solution

Answer

The partial fraction decomposition of the given rational expression \(\frac{x}{(x+1)^{2}}\) is 0.

Step 1: Identify the form and degree of the expression

In this exercise, the given expression is \(\frac{x}{(x+1)^{2}}\). This is a rational function, where the numerator is of degree 1 (x) and the denominator is of degree 2 \((x+1)^{2}\). The degree of the numerator is less than that of the denominator. Therefore, the expression is a proper fraction.

Step 2: Express the Denominator as Product of Linear Factors

The denominator of the fraction \((x+1)^{2}\) is already factored into linear parts. Each of the linear factors is \((x+1)\).

Step 3: Set up the Partial Fraction Decomposition

The next step is to set up the partial fraction decomposition. Each of the unique linear factor gives rise to a term in the decomposition. As the factor \((x+1)\) is repeated twice, there will be a term for \((x+1)\) and for \((x+1)^{2}\). The unknown coefficients need to be determined through later steps - \(\frac{x}{(x+1)^2} = \frac{A}{x+1} + \frac{B}{(x+1)^2}\).

Step 4: Find the Unknown Coefficients

To find the unknown coefficients, multiply throughout by the common denominator \((x+1)^2\), thereby getting rid of the fractions, which results in the equation: \(x = A*(x+1) + B\). This equation must hold true for all values of x. Choose particular handy values of x to solve for A and B. A suitable choice here is x = -1, directly giving us the value for B. Plug in x = -1, the equation simplifies to: 0 = A*0 + B, which imply B = 0. Since we know the value of B, choose another x value, say x = 0, to solve for A. Plugging in x = 0, gets: 0 = A*1 + 0, which means A = 0.

Step 5: Write the final expression

Substituting the values for A and B obtained in the previous step, we have: \(\frac{x}{(x+1)^{2}} = \frac{0}{x+1} + \frac{0}{(x+1)^{2}} ~= 0 + 0 = 0\)

Key Concepts

Rational Expressions

A key concept in algebra is that of rational expressions, which are essentially fractions where the numerator and denominator are both polynomials. For example, \( \frac{x}{(x+1)^{2}} \) is a rational expression, with \( x \) being the numerator polynomial, and \( (x+1)^{2} \) as the denominator polynomial.

The goal with these expressions is often to simplify or rewrite them in a way that makes them easier to work with, especially when integrating or finding limits in calculus. Partial fraction decomposition is one such technique used to break down more complex rational expressions into simpler, more manageable pieces. It's particularly helpful when you're faced with integral or differential equations.

In algebra classes, understanding and being able to work with rational expressions is essential, as they're common in calculus and higher-level math courses. Managing these expressions includes simplifying them, finding their domains (since the denominator cannot be zero), and rewriting them when necessary.

Linear Factors

Linear factors are an important component when working with polynomial expressions. The term 'linear' refers to the polynomial being of the first degree, which means it can be written in the form \( ax + b \) where \( a \) and \( b \) are constants, and \( x \) is the variable.

In partial fraction decomposition, a key step involves factoring the denominator of the rational expression into its linear (or sometimes irreducible quadratic) factors. For instance, in the example exercise \( (x+1)^2 \) is the squared linear factor of our denominator. The significance here is that each of these factors will correspond to a part of the decomposed form.

Why focus on linear factors?

Linear factors are the building blocks for polynomial expressions. In many mathematical situations, breaking down a complex polynomial into its linear components simplifies the problem and provides a clearer path to the solution. This is evident in solving polynomial equations, where finding the roots involves factoring to the most basic linear components.

Unknown Coefficients

When it comes to partial fraction decomposition, one of the crucial steps is determining the unknown coefficients in the decomposed terms. These coefficients are essentially placeholders that we solve for so that the decomposed expression will be equivalent to the original fraction.

Using the provided example \( \frac{x}{(x+1)^2} \) becoming \( \frac{A}{x+1} + \frac{B}{(x+1)^2} \), \( A \) and \( B \) are the unknown coefficients which we need to find to complete the decomposition. To do this, we set up an equation that must hold for all values of \( x \) by multiplying through by the common denominator, thereby eliminating the fractions.

Techniques for Finding Coefficients

There are different methods to find these values, such as the equation solving approach used in the example or polynomial matching. Generally, setting up and solving a system of equations based on strategically chosen values of \( x \) can quickly yield these unknowns. In particular, values of \( x \) that can simplify the equations are chosen, such as the roots of the denominators of the original expression. These steps are fundamental in ensuring that the partial fraction decomposition is correctly performed and that the original rational expression is accurately represented.