Problem 40 - Precalculus with Limits

Question

Writing the Partial Fraction Decomposition. Write the partial fraction decomposition of the rational expression. Check your result algebraically. $$\frac{x^{2}-4 x+7}{(x+1)\left(x^{2}-2 x+3\right)}$$

Step-by-Step Solution

Answer

The partial fraction decomposition is found by setting the given equation: \( \frac{x^2 - 4x + 7}{(x+1)(x^2 - 2x + 3)} \) equal to \( \frac{A}{x + 1} + \frac{Bx + C}{x^2 - 2x + 3} \), and solving for constants A, B and C. It is checked by simplifying the partial sum and ensuring that it equals the original expression.

Step 1: Identify Factors

First, factorize the denominator part of the expression, if possible. The factors for the denominator \( (x+1)(x^2-2x+3) \) are \(x+1\) and \(x^2 - 2x + 3\). These factors do not further factorize, so they will be used as the denominators for the partial fraction decomposition.

Step 2: Decompose into Partial Fractions

Set the expression \( \frac{x^2 - 4x + 7}{(x+1)(x^2 - 2x + 3)} \) equal to the sum of two fractions, \( \frac{A}{x + 1} \) and \( \frac{Bx + C}{x^2 - 2x + 3} \) where A, B and C are constants. Therefore, the equation becomes \( x^2 - 4x + 7 = A(x^2 - 2x + 3) + (Bx + C)(x + 1) \). This equation must be true for all values of \(x\).

Step 3: Solve for Constants A,B,C

To calculate the constants A, B and C, select easy plug-in values for \(x\), typically 0 or 1. Then convert the equation to a system of linear equations and solve for A, B and C. After calculating these values, plug them back into the partial fractions to check if the original equation holds.

Step 4: Cross Check the Result

To check if the solution is correct, perform the fraction addition operation and simplify the equation. If it gives the original equation, then the partial fraction decomposition is correct.

Key Concepts

Rational Expressions

Rational expressions are fractions wherein both the numerator and the denominator are polynomials. Just like numerical fractions, they represent a division of two quantities. Their most essential properties include being able to simplify by factoring, finding restrictions for the variable (since the denominator cannot be zero), and performing operations such as addition, subtraction, multiplication, and division.

To understand partial fraction decomposition, imagine breaking a complex rational expression into simpler fractions that are easier to work with. This method is particularly useful in calculus for integration and solving differential equations. The key to successfully decomposing a rational expression lies in the correct factorization of the denominator and numerically solving for the unknowns.

Factorizing Polynomials

Factorizing polynomials is a critical step in the process of partial fraction decomposition. The goal of factorization is to rewrite a polynomial as the product of its simplest polynomials. There are various methods for factorizing, including finding the greatest common factor, using the difference of squares, sum and difference of cubes, and the quadratic formula for trinomials.

For the partial fraction decomposition exercise given, the denominator, \( (x+1)(x^2-2x+3) \), is already factored and cannot be simplified further. Recognizing that each factor is unique and unable to break down anymore is crucial for setting up the correct form for the partial fractions. Without proper factorization, it is impossible to proceed with the decomposition.

System of Linear Equations

The final step of partial fraction decomposition is solving a system of linear equations. These are a collection of linear equations with the same variables, which in the context of partial fraction decomposition are the coefficients of the decomposed fractions. Solving these systems can be done using substitution, elimination, or matrix-related methods.

After decomposing the original fraction and aligning terms, we set coefficients of like terms on both sides of the equation equal to one another. This gives us several linear equations to solve concurrently. Finding a solution to this system offers the values for the unknown constants in the decomposed fractions, completing the decomposition process.