Problem 9 - Precalculus with Limits

Question

Writing the Form of the Decomposition. Write the form of the partial fraction decomposition of the rational expression. Do not solve for the constants. $$\frac{3}{x^{2}-2 x}$$

Step-by-Step Solution

Answer

The general form of the partial fraction decomposition is \(\frac{A}{x} + \(\frac{B}{x-2}\).

Step 1: Identify the denominator

First identify the denominator of the rational function, which in this case is \(x^{2}-2x\). The next step is to factor this polynomial.

Step 2: Factor the denominator

Factor the polynomial in the denominator. In this case, \(x^{2}-2x\) can be factored as \(x*(x-2)\). So our denominator has two distinct linear factors, \(x\) and \(x-2\).

Step 3: Write the general form of the partial fraction decomposition

For each linear factor \(ax+b\) in the denominator of the rational function, there's a corresponding term \(\frac{A}{ax+b}\) in the decomposition. Because our denominator has two distinct linear factors \(x\) and \(x-2\), the general form of the partial fraction decomposition should be \(\frac{A}{x} + \(\frac{B}{x-2}\), where A and B are constants to be determined.

Key Concepts

Rational Expressions

A rational expression is essentially a fraction where both the numerator and the denominator are polynomials. These expressions can represent relationships or describe various mathematical scenarios. When dealing with rational expressions, it’s crucial to ensure that the denominator is not equal to zero, which would make the expression undefined.
Consider the expression \(\frac{3}{x^{2}-2x}\). This is a typical rational expression because the numerator is a constant, 3, and the denominator is a polynomial, \(x^{2}-2x\). Rational expressions can be simplified, evaluated, or decomposed into simpler parts when needed. Understanding and working with these expressions can make solving algebraic equations easier, especially when you're performing operations like addition, subtraction, or division.

Linear Factors

Linear factors are polynomial expressions of the first degree, which means they can be expressed in the form \(ax+b\), where \(a\) and \(b\) are constants.
When expressions like \(x^2-2x\) are factorized, they reveal their linear components, which are essential in solving and simplifying algebraic equations.
In our example, the factorization of the denominator \(x^2-2x\) yields two linear factors: \(x\) and \(x-2\). These factors form the backbone of partial fraction decomposition, especially because they directly indicate how the rational expression breaks down into simpler fractions.

Factoring Polynomials

To factor a polynomial is to express it as a product of other polynomials. This often involves recognizing patterns or applying techniques like the factor theorem, grouping, or special identities.
In our case, the denominator \(x^2-2x\) is factored by taking out the greatest common factor, \(x\). This results in \(x(x-2)\).
Steps to factor:

• Identify the Greatest Common Factor (GCF): Start by checking if there is a common factor in each term. Here, \(x\) is common in both terms, \(x^2\) and \(2x\).
• Rewrite the polynomial: By factoring out \(x\), rewrite \(x^2-2x\) as \(x(x-2)\).

Factoring is vital for simplifying rational expressions and is an essential step in partial fraction decomposition.

Mathematical Constants

In a partial fraction decomposition, mathematical constants are the unknowns represented by letters such as \(A\) and \(B\). These constants are crucial as they define the coefficients in the simplified fractional parts of the decomposition.
These constants will be determined by solving for specific values that satisfy the original rational expression. Though they aren't solved immediately in this partial fraction setup, recognizing them is essential. They serve to balance the equation, ensuring that the decomposition faithfully replicates the original expression.
The role of constants like \(A\) and \(B\) is to maintain the equivalence between the decomposed form of the rational expression and its initial form. Although they seem elusive at first, they are pivotal in the final solution of such problems.