Problem 1 - 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. \(\frac{1}{(x-1)(x+2)}\)

Step-by-Step Solution

Answer

\( \frac{1}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2} \)

Step 1: Identify the Denominator Factors

The first step is to identify and list the distinct linear factors found in the denominator of the given fraction. For this problem, the denominator \((x-1)(x+2)\) has two distinct factors: \((x-1)\) and \((x+2)\).

Step 2: Write the General Form of the Decomposition

Using the identified factors, write the general form of the partial fraction decomposition. Since both factors are linear, the form will include terms with unknown coefficients, \(A\) and \(B\), each over their respective factors. The decomposition looks like: \[ \frac{1}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2} \].

Key Concepts

Polynomial Denominator

In partial fraction decomposition, the first important step is to understand the polynomial denominator of the given fraction. A polynomial, in simple terms, is a mathematical expression consisting of variables and coefficients. It can have terms that are constants, variables, or products of a constant and variable raised to an exponent. In this particular function, we examine the denominator \[ (x-1)(x+2) \]. This expression is a polynomial made by multiplying two linear binomials. Identifying these terms is crucial, as each distinct factor in the polynomial determines a separate term in the decomposition. Recognizing these will guide us in rewriting the fraction into simpler terms that are easier to integrate or differentiate.

Linear Factors

To perform a partial fraction decomposition, identifying linear factors is essential. Linear factors are components of a polynomial that have an exponent of one, making them straight-line equations when graphed. In our example, the polynomial denominator \( (x-1)(x+2) \) consists of two distinct linear factors:

• \( (x-1) \)
• \( (x+2) \)

Linear factors are crucial because they influence how we break down the fraction. Each linear factor leads to a distinct term in the partial fraction, with its unknown coefficient which needs to be identified. We will use these linear factors to express the original fraction as a sum of simpler fractions, each involving one of the linear factors in the denominator.

Coefficient Identification

After recognizing the polynomial's linear factors, the next step is to set up the structure for coefficient identification. In the partial fraction decomposition, we translate the original fraction based on these linear factors and assign each term an unknown coefficient. In our case, the setup will be:\[\frac{1}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2}\]Here, \(A\) and \(B\) are the unknown coefficients we need to find. The equation's left side equals the right side, but each linear factor is isolated with its coefficient. Determining \(A\) and \(B\) typically involves substituting values into the equation or equating coefficients from both sides of the equation after finding common denominators. Instead of solving for these here, noting their presence helps prepare for calculations that often occur in further mathematical work, making complex functions much simpler to handle.