Problem 4 - College Algebra

Question

Write the form of the partial fraction decomposition of the rational expression. It is not necessary to solve for the constants. $$\frac{3 x+16}{(x+1)(x-2)^{2}}$$

Step-by-Step Solution

Answer

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

Step 1: Identify the factors in the denominator

The given rational function has the denominator (x + 1)(x - 2)^2. It consists of a linear factor (x + 1) and a repeated linear factor (x - 2)^2.

Step 2: Write the general form of partial fraction decomposition

The general form of partial fraction decomposition for linear and repeated linear factors is \(\frac{A}{(x + 1)} + \frac{B}{(x - 2)} + \frac{C}{(x - 2)^2}\). Here, A, B and C are constants that we would solve for if we needed the complete decomposition.

Step 3: Complete the partial fraction decomposition

Without knowing A, B and C, our partial fraction decomposition will remain in this form. Hence, the decomposition of the given rational function is \(\frac{A}{(x + 1)} + \frac{B}{(x - 2)} + \frac{C}{(x - 2)^2}\).

Key Concepts

Rational Expressions

Rational expressions are fractions where the numerator and the denominator are both polynomials. In the case of the given exercise, \( \frac{3x+16}{(x+1)(x-2)^2} \) is a rational expression because both 3x+16 (the numerator) and \( (x+1)(x-2)^2 \) (the denominator) are polynomials.

The study of rational expressions is fundamental in algebra, as they appear in many different contexts, from simplifying algebraic fractions to solving equations. Understanding how to handle these expressions, including how to perform operations like addition, subtraction, multiplication, division, and simplification, is key to working with them effectively.

Just like numerical fractions, rational expressions can often be simplified by factoring and reducing common factors. But when it comes to division, things get a bit more intricate, and this is where partial fraction decomposition comes into play. This technique breaks down complex rational expressions into simpler, more manageable pieces, especially useful for integration in calculus.

Linear Factors

Linear factors refer to expressions of the form \( ax + b \), where \( a \) and \( b \) are constants and \( a \) is not zero. They are 'linear' because they represent straight lines when plotted on a graph. The denominator of the rational expression from the exercise, \( (x+1)(x-2)^2 \) contains one such linear factor, \( x+1 \).

In the context of partial fraction decomposition, each distinct linear factor in the denominator will correspond to a separate term in the decomposition. So, if we only had distinct linear factors, our partial fraction would look something like \( \frac{A}{x+1} + \frac{B}{x-2} + \cdots \), with \( A \) and \( B \) being constants that could later be determined.

Repeated Linear Factors

Repeated linear factors occur when a linear factor is raised to a power greater than one, such as \( (x-2)^2 \) in the given exercise's denominator. In partial fraction decomposition, each occurrence of the repeated factor must be accounted for separately.

For the factor \( (x-2)^2 \), this means we need to include terms of the form \( \frac{B}{x-2} \) and \( \frac{C}{(x-2)^2} \) in our decomposition, where \( B \) and \( C \) are constants. Each power of the repeated factor, up to the highest power present in the denominator, should have its own term in the decomposition. This step ensures that, when combined, these terms can represent any possible numerator that could pair with the repeated linear factor in the original rational expression.

Algebraic Fractions

Algebraic fractions are simply fractions with algebraic expressions in the numerator and/or denominator—similar to rational expressions but not limited to polynomials. Partial fraction decomposition is a technique often used to simplify algebraic fractions, making them easier to work with.

In calculus, for example, algebraic fractions often occur in integration problems. Decomposing a complex algebraic fraction into simpler fractions makes finding the integral much more straightforward. Going through the partial fraction decomposition process, as shown in the exercise solution, transforms a complicated algebraic fraction into a sum of simpler fractions, where standard integration techniques can be applied to each term individually.