Problem 29 - Algebra and Trigonometry

Question

write the partial fraction decomposition of each rational expression. $$\frac{5 x^{2}-6 x+7}{(x-1)\left(x^{2}+1\right)}$$

Step-by-Step Solution

Answer

The partial fraction decomposition of the rational expression \( \frac{5 x^{2}-6 x+7}{(x-1)(x^{2}+1)}\) is \( \frac{5}{x - 1} - \frac{5x + 1}{x^2 + 1}\).

Step 1: Identify the denominator factors

In this rational expression, the denominator has two factors: \(x - 1\) and \(x^2 + 1\).

Step 2: Form the general form of the partial fractions

Given the factors in the denominator, the general form of the partial fraction decomposition is: \( \frac{5 x^{2}-6 x+7}{(x-1)(x^2 +1)} = \frac{A}{x - 1} + \frac{Bx + C}{x^2 + 1} \)

Step 3: Simplify the right hand side of the equation to common fractions

To find the constants A, B, and C, write the right side as a single fraction. This gives \((Bx + C) (x - 1) + A (x^2 + 1) = 5x^2 - 6x + 7\)

Step 4: Equate coefficients of the same powers of x on both sides

This gives us: \(A + B = 0 \), (from coefficient of \(x^2\)), \(- B + C = -6\) (from coefficient of \(x\)), and \(-C + A = 7\) (from the constant term).

Step 5: Solve for A, B, and C

The system of linear equations can then be solved to obtain the coefficients A, B, and C. Solving it gives : \(A = 5, B = -5\) and \(C = -1\).

Step 6: Substitute A, B, C values into partial fractions

Substitute the values of A, B, and C into the general form of the partial fractions. This gives the final partial fraction decomposition as \( \frac{5}{x - 1} - \frac{5x + 1}{x^2 + 1}\).

Key Concepts

Rational Expressions

Rational expressions are like fractions but instead of just numbers in the numerator and denominator, you might find polynomials. A rational expression is any expression that can be written in the form \(\frac{P(x)}{Q(x)}\), where \(P(x)\) and \(Q(x)\) are polynomials and \(Q(x) eq 0\). These expressions appear frequently in algebra. Understanding them is crucial for solving complex algebraic equations. When dealing with rational expressions, it's important to simplify them by factoring and cancelling common factors.

In the case of the given exercise, the rational expression \( \frac{5x^2 - 6x + 7}{(x-1)(x^2 +1)} \) has a polynomial in the numerator and a polynomial in the denominator. This complexity necessitates a more advanced technique, like partial fraction decomposition, to break it down into simpler components. This method involves writing a complex rational expression as the sum of simpler fractions, making it easier to work with in algebraic operations, such as integration in calculus or solving algebraic equations.

Algebra

Algebra is the branch of mathematics dealing with symbols and rules for manipulating those symbols. It provides the tools for solving equations and understanding how those equations relate to one another. In this specific situation, we're using algebra to decompose a rational expression into partial fractions.

To decompose a fraction, you first have to understand the algebraic structure of the denominator. Identify the factors, which are the building blocks of the expression. For this exercise, we identified the denominator \((x-1)(x^2 + 1)\) as having two factors. These factors guide how we split the rational expression. Using principles of algebra, we set equations according to the different powers and solved them to find unknowns represented by constants like \(A, B,\) and \(C\).

The partial fraction decomposition exploits the ability to re-express a single complex expression as a sum or combination of simpler rational expressions, thus highlighting the fundamental connectivity between complex equations and basic algebraic tools.

Linear Equations

Linear equations are equations of the first degree. They're incredibly useful for solving problems involving unknowns. In the partial fraction decomposition process, we encounter a set of linear equations when we match coefficients of like terms from both sides of the equation.

For this exercise to solve for \( A, B,\) and \( C \), linear equations were formed from matching like powers of \( x \). The system of equations obtained was:
\[ A + B = 0 \ -B + C = -6 \ -C + A = 7 \]
Linear equations such as these can be solved using various algebraic methods, but in this case, basic substitution or elimination methods suffice. The solutions \(A = 5\), \(B = -5\), and \(C = -1\) were found using these techniques. This solution shows how linear equations can help break down complex expressions into components that are easier to work with and interpret.