Problem 11 - Algebra for College Students

Question

Find the partial fraction decomposition for each rational expression. See answers below. \(\frac{2 x+1}{(x-2)^{2}}\)

Step-by-Step Solution

Answer

\( \frac{2}{x-2} + \frac{5}{(x-2)^2} \)

Step 1: Identify the Form of Decomposition

The given rational expression is \( \frac{2x + 1}{(x-2)^2} \). Since \((x-2)^2\) is a repeated linear factor, the partial fraction decomposition will have the form \( \frac{A}{x-2} + \frac{B}{(x-2)^2} \), where \( A \) and \( B \) are constants to be determined.

Step 2: Assume Partial Fraction and Clear Denominators

Write the assumed partial fraction decomposition: \( \frac{2x + 1}{(x-2)^2} = \frac{A}{x-2} + \frac{B}{(x-2)^2} \). Multiply through by \((x-2)^2\) to eliminate the denominators, obtaining: \( 2x + 1 = A(x-2) + B \).

Step 3: Expand and Simplify the Equation

Expand \( A(x-2) \) to get \( Ax - 2A \). The equation now reads: \( 2x + 1 = Ax - 2A + B \).

Step 4: Equate and Solve for Constants

Equate coefficients of like terms from both sides of the equation:\1. Coefficient of \( x \): \( 2 = A \)\2. Constant terms: \( 1 = -2A + B \).Using \( A = 2 \), substitute into the second equation: \( 1 = -2(2) + B \), resulting in \( 1 = -4 + B \), thus \( B = 5 \).

Step 5: Write the Final Decomposition

Substitute \( A = 2 \) and \( B = 5 \) back into the partial fraction form: \( \frac{2x + 1}{(x-2)^2} = \frac{2}{x-2} + \frac{5}{(x-2)^2} \).

Key Concepts

Rational Expressions

Rational expressions are fractions that have polynomials in their numerator and denominator. They can be quite complex to work with, especially when simplifying or decomposing them. The expression \( \frac{2x + 1}{(x - 2)^2} \) is an example of a rational expression. Here, the numerator is \( 2x + 1 \) and the denominator is \( (x - 2)^2 \). The goal in partial fraction decomposition is to break down a complex rational expression into a sum of simpler fractions. This helps in understanding the behavior of the expression, especially when integrating or differentiating. It's much like breaking down complex numbers into basic units. With rational expressions, always ensure that the degree of the polynomial in the numerator is less than the degree of the polynomial in the denominator before proceeding with decomposition. If not, divide the numerator by the denominator first to simplify.

Repeated Linear Factors

In polynomial algebra, repeated linear factors are roots that appear more than once in a polynomial's factorization. For example, in the expression \((x-2)^2\), the factor \(x-2\) is repeated twice. This characteristic influences how the partial fraction decomposition is structured. When dealing with repeated linear factors, each power of the factor must be accounted for in the decomposition. For \((x-2)^2\), this means including terms for both \(\frac{A}{x-2}\) and \(\frac{B}{(x-2)^2}\). This ensures that all potential partial fractions are evaluated, helping in achieving an accurate decomposition. Handling repeated factors might feel challenging, but recognizing them early simplifies the process.

Constants Determination

Once the form of the partial fraction decomposition is set, the next step is to determine the constants. This involves substituting the partial fraction form back into the original equation and eliminating the denominators by multiplying. This results in an equation that can be expanded and simplified to aid in solving for the constants. In the given problem, this involves multiplying \(2x + 1 = A(x-2) + B\), allowing for expansion and rearrangement to separate terms based on degree. To find the constants \(A\) and \(B\), equate the coefficients of corresponding terms on either side of the equation:

• Coefficient of \(x\): This gives \(2 = A\).
• Constant terms: Solving \(1 = -2A + B\) using \(A = 2\) results in \(B = 5\).

Hence, the constants are integral in ensuring the partial fraction form accurately represents the original rational expression. Properly determining them is crucial for correct decomposition.