Problem 38 - Precalculus: Mathematics for Calculus

Question

Find the partial fraction decomposition of the rational function. $$\frac{3 x^{2}-2 x+8}{x^{3}-x^{2}+2 x-2}$$

Step-by-Step Solution

Answer

\( \frac{4}{x - 1} + \frac{-x + 1}{x^2 + 2} \)

Step 1: Identify the Denominator Polynomial

The first step in finding the partial fraction decomposition is identifying the polynomial in the denominator. Here, we have \( x^3 - x^2 + 2x - 2 \).

Step 2: Factor the Denominator Polynomial

Try to factor the denominator polynomial \( x^3 - x^2 + 2x - 2 \). You can start by checking for possible rational roots using the Rational Root Theorem.After testing possible roots, we find that \( x = 1 \) is a root. So, divide \( x^3 - x^2 + 2x - 2 \) by \( x - 1 \) using synthetic or polynomial division.After dividing, we get \( x^3 - x^2 + 2x - 2 = (x - 1)(x^2 + 2) \).

Step 3: Write the Partial Fractions

Since the denominator can be factored as \( (x - 1)(x^2 + 2) \), set up the partial fraction decomposition:\[ \frac{3x^2 - 2x + 8}{(x - 1)(x^2 + 2)} = \frac{A}{x - 1} + \frac{Bx + C}{x^2 + 2} \]

Step 4: Multiply Through by the Denominator

Multiply the entire equation by the original denominator, \( (x - 1)(x^2 + 2) \), to clear the fractions:\[ 3x^2 - 2x + 8 = A(x^2 + 2) + (Bx + C)(x - 1) \]

Step 5: Expand and Collect Like Terms

Expand the right-hand side:\[ A(x^2 + 2) = Ax^2 + 2A \]\[ (Bx + C)(x - 1) = Bx^2 - Bx + Cx - C \]Combine all terms:\[ Ax^2 + 2A + Bx^2 - Bx + Cx - C = (A + B)x^2 + (C - B)x + (2A - C) \]

Step 6: Determine the Coefficients

Match coefficients from both sides of the equation:- For \(x^2\): \( A + B = 3 \)- For \(x\): \( C - B = -2 \)- Constant term: \( 2A - C = 8 \)Solve this system of equations:1. \( A + B = 3 \)2. \( C - B = -2 \)3. \( 2A - C = 8 \)From these, we find: \( A = 4, B = -1, C = 1 \).

Step 7: Write the Final Decomposition

Substitute the values of \( A \), \( B \), and \( C \) back into the partial fraction decomposition:\[ \frac{4}{x - 1} + \frac{-x + 1}{x^2 + 2} \]This is the partial fraction decomposition.

Key Concepts

Factor the Denominator Polynomial

Factoring the denominator polynomial is an essential skill in partial fraction decomposition. Here, we start with the polynomial \( x^3 - x^2 + 2x - 2 \). The goal is to express this polynomial as a product of simpler polynomials.

Perform the following steps:

• Use the Rational Root Theorem to find possible roots. The theorem suggests possible rational roots are factors of the constant term, \(-2\), divided by factors of the leading coefficient, \(1\). Hence, the potential roots are \( \pm 1 \) and \( \pm 2\).
• Test each possible root using substitution or synthetic division. By substitution, you find that \( x = 1 \) is a root.
• Now, divide the polynomial by \( x - 1 \) using synthetic or polynomial division. This will simplify \( x^3 - x^2 + 2x - 2 \) to \((x - 1)(x^2 + 2)\).

The process concludes by expressing the polynomial in factored form: \((x - 1)(x^2 + 2)\). This is a crucial step, simplifying the process for partial fraction decomposition.

Rational Function

Rational functions are expressions that involve the division of two polynomials. In our example, we have the rational function \( \frac{3x^2 - 2x + 8}{x^3 - x^2 + 2x - 2} \).

Key points to understand about rational functions:

• The numerator is the polynomial on the top of a fraction, here it's \(3x^2 - 2x + 8\). It determines the contribution of the function's values for different \(x\).
• The denominator, \(x^3 - x^2 + 2x - 2\), should not equal zero, as it defines where the function is undefined. Because of its higher degree compared to the numerator, this rational function is considered "proper," stabilizing as \(x\) approaches extreme values.
• Rational functions can model different behaviors such as steady growth, decline, or irregular oscillation depending on their coefficients and the roots of the denominator. Factoring the denominator shows us potential vertical asymptotes and simplifications for partial fraction decomposition.

Grasping these fundamentals will help you analyze and manipulate rational functions effectively in various contexts.

Solve System of Equations

Solving a system of equations is often necessary after setting up the partial fraction decomposition. In our exercise, once the equation is multiplied out, we compare coefficients to form a system:

• \( A + B = 3 \)
• \( C - B = -2 \)
• \( 2A - C = 8 \)

To solve this system:

• First, rearrange and substitute between equations to isolate each variable. For example, express \(A\) from the first equation as \(A = 3 - B\).
• Substitute \(A\) in the third equation: \(2(3 - B) - C = 8\). Simplify to find \(C = 2B - 2\).
• Replace \(C\) in the second equation: \(2B - 2 - B = -2\). Solve to find \(B = -1\).
• Plug \(B\) back into expressions for \(A\) and \(C\) to find \(A = 4\) and \(C = 1\).

Understanding these relationships forms the bedrock of decomposing complex rational expressions. Once you've mastered these steps, similar systems of equations will be manageable in future scenarios.