Problem 40 - Precalculus

Question

Find the partial-fraction decomposition for each rational function. $$\frac{5 x+2}{x^{3}-8}$$

Step-by-Step Solution

Answer

\(\frac{5x + 2}{x^3-8} = \frac{1}{x-2} + \frac{-x+1}{x^2+2x+4}\).

Step 1: Factor the Denominator

First, we need to factor the denominator of the rational function \(x^3 - 8\). We can recognize \(x^3 - 8\) as a difference of cubes, \(a^3 - b^3\), which factors into \((a-b)(a^2 + ab + b^2)\). Thus, \(x^3 - 8 = (x-2)(x^2 + 2x + 4)\).

Step 2: Set Up the Partial Fraction Form

Since \(x^3 - 8\) factors into \((x-2)(x^2 + 2x + 4)\), we write the partial fraction decomposition as: \( \frac{5x+2}{x^3-8} = \frac{A}{x-2} + \frac{Bx+C}{x^2+2x+4} \). This represents the decomposed form where \(A, B,\) and \(C\) are constants we need to find.

Step 3: Clear the Denominator

To clear the denominators, multiply both sides of the equation by \((x-2)(x^2 + 2x + 4)\) resulting in \(5x + 2 = A(x^2 + 2x + 4) + (Bx + C)(x-2)\).

Step 4: Expand Right-Hand Side

Expand the right-hand side: \[A(x^2 + 2x + 4) + (Bx + C)(x-2) = Ax^2 + 2Ax + 4A + Bx^2 - 2Bx + Cx - 2C.\] Combine like terms to get: \[(A + B)x^2 + (2A - 2B + C)x + (4A - 2C).\]

Step 5: Equate Coefficients

Match coefficients of \(x^2\), \(x\), and the constant terms from both sides of the equation. This gives us the system of equations:1. \(A + B = 0\)2. \(2A - 2B + C = 5\)3. \(4A - 2C = 2\).

Step 6: Solve for Constants

Solve the system of equations obtained in the previous step:From equation 1: \(B = -A\).Substitute \(B = -A\) in equation 2:\(2A - 2(-A) + C = 5 \Rightarrow 4A + C = 5\). (Equation 4)From equation 3:\(4A - 2C = 2\).Now, solve equations 4 and 3 together:- Equation 4: \(4A + C = 5\)- Equation 3: \(4A - 2C = 2\)From equation 3, solve for \(C\):\(4A - 2C = 2 \Rightarrow -2C = 2 - 4A \Rightarrow C = 2A - 1\).Substitute \(C = 2A - 1\) in equation 4:\(4A + (2A - 1) = 5 \Rightarrow 6A - 1 = 5 \Rightarrow 6A = 6 \Rightarrow A = 1\).Now \(C = 2(1) - 1 = 1\) and from \(B = -A\), \(B = -1\).Thus, \(A = 1\), \(B = -1\), \(C = 1\).

Step 7: Write the Final Decomposition

Substitute \(A = 1\), \(B = -1\), and \(C = 1\) back into the partial fraction form: \[\frac{5x + 2}{x^3-8} = \frac{1}{x-2} + \frac{-x+1}{x^2+2x+4}\].

Key Concepts

Rational Functions

Rational functions are expressions that represent the ratio of two polynomials. In mathematical terms, a rational function is any function that can be written as \( \frac{P(x)}{Q(x)} \), where \( P(x) \) and \( Q(x) \) are polynomials and \( Q(x) \) is non-zero. These functions can sometimes produce complex fractions that we simplify using various techniques.

For example, consider the function \( \frac{5x+2}{x^3-8} \). The polynomial \( 5x + 2 \) is the numerator, and \( x^3 - 8 \) is the denominator. To simplify such expressions, we often use methods like factoring or partial fraction decomposition. This helps us break down complex rational expressions into simpler, more manageable parts. Simplifying rational functions is a crucial skill, especially in calculus, where they often appear in integrals and limits.

Difference of Cubes

The difference of cubes is a specific type of polynomial expression that takes the form \( a^3 - b^3 \). These types of expressions have a well-known factorization formula: \( a^3 - b^3 = (a-b)(a^2 + ab + b^2) \). This formula allows us to factor the expression into a product of two factors, simplifying further calculations.

For instance, the expression \( x^3 - 8 \) can be identified as a difference of cubes because it resembles the format \( x^3 - 2^3 \). Applying the formula, we get \( (x - 2)(x^2 + 2x + 4) \). This factoring is essential as it helps to rewrite rational functions in a more usable form for methods like partial fraction decomposition. Gaining comfort with factoring differences of cubes can simplify many algebraic problems.

Equating Coefficients

Equating coefficients is a technique used to identify unknown constants in polynomial equations. When we break down the original expression into a sum of simpler fractions, we match the coefficients of corresponding terms from both sides to derive a system of equations.

In the case of the rational function \( \frac{5x + 2}{x^3 - 8} \), after setting up the partial fraction decomposition, we arrived at an equation where the expanded terms of both sides need to match: \( 5x + 2 \) equals the expanded expression of \(A(x^2 + 2x + 4) + (Bx + C)(x - 2)\). By aligning the coefficients of like powers of \( x \), we obtain equations for the constants \( A \), \( B \), and \( C \). For instance, matching the coefficients gives:

• \( A + B = 0 \)
• \( 2A - 2B + C = 5 \)
• \( 4A - 2C = 2 \)

Solving these equations helps determine the values of the constants, thus completing the partial fraction decomposition process.

System of Equations

A system of equations involves multiple equations being solved simultaneously. Each equation within the system shares common variables that we need to solve for. System of equations are often employed to find solutions to unknowns in processes like partial fraction decomposition.

For the expression \( \frac{5x + 2}{x^3-8} \), once we equate coefficients, we derive a set of linear equations involving constants \( A \), \( B \), and \( C \):

• \( A + B = 0 \)
• \( 2A - 2B + C = 5 \)
• \( 4A - 2C = 2 \)

This system needs to be solved to find the specific values of \( A \), \( B \), and \( C \). By using methods like substitution or elimination, we solve the equations step by step. This allows us to substitute back into our partial fraction model, finalizing the decomposition. Understanding how to navigate and solve such systems is a foundational algebraic skill, critical for advanced mathematical problem-solving.