Problem 38 - College Algebra

Question

Find the decomposition of the partial fraction for the irreducible non repeating quadratic factor. \(\frac{-5 x^{2}+18 x-4}{x^{3}+8}\)

Step-by-Step Solution

Answer

The partial fraction decomposition is \( \frac{-5x + 22}{x^2 - 2x + 4} \).

Step 1: Factor the Denominator

First, recognize that the denominator, \(x^3 + 8\), can be factored as \((x+2)(x^2 - 2x + 4)\). This utilizes the sum of cubes: \(a^3 + b^3 = (a+b)(a^2 - ab + b^2)\).

Step 2: Set Up the Partial Fraction

Since \(x^3 + 8 = (x+2)(x^2 - 2x + 4)\), express the rational function as a sum of two fractions: \[ \frac{-5x^2 + 18x - 4}{x^3 + 8} = \frac{A}{x+2} + \frac{Bx + C}{x^2 - 2x + 4}. \]

Step 3: Clear the Denominator

Multiply the entire equation by the original denominator \((x^3 + 8)\) to clear the fractions. This results in: \[-5x^2 + 18x - 4 = A(x^2 - 2x + 4) + (Bx + C)(x+2).\]

Step 4: Expand and Combine Like Terms

Expand the expressions on the right-hand side: \[A(x^2 - 2x + 4) = Ax^2 - 2Ax + 4A,\] \[(Bx + C)(x+2) = Bx^2 + 2Bx + Cx + 2C.\] Combine and arrange the terms: \[(A + B)x^2 + (2B - 2A + C)x + (4A + 2C).\]

Step 5: Set Up Equations to Solve for Constants

The equation \(-5x^2 + 18x - 4 = (A + B)x^2 + (2B - 2A + C)x + (4A + 2C)\) provides a system of equations:1. \(A + B = -5\)2. \(2B - 2A + C = 18\)3. \(4A + 2C = -4\)

Step 6: Solve the System of Equations

Solve the equations from Step 5:1. From \(A + B = -5\), express \(A\) as \(A = -5 - B\).2. Substitute \(A = -5 - B\) into equations 2 and 3, solve for \(B\) and \(C\).3. Solving gives \(A = 0, B = -5, C = 22\).

Step 7: Write the Final Partial Fraction Decomposition

Substituting the constants back into the partial fraction form gives: \[\frac{A}{x+2} + \frac{Bx + C}{x^2 - 2x + 4} = \frac{0}{x+2} + \frac{-5x + 22}{x^2 - 2x + 4}.\] Therefore, the decomposition simplifies to \(\frac{-5x + 22}{x^2 - 2x + 4}.\)

Key Concepts

Sum of Cubes

The sum of cubes is a mathematical identity used to factor the sum of two cubed terms. When you encounter an expression like \(a^3 + b^3\), it can be factored into

• \(a^3 + b^3 = (a + b)(a^2 - ab + b^2)\)

This identity simplifies complex polynomials, making them easier to handle. For instance, in the original exercise, we factored \(x^3 + 8\) using this identity, where \(x^3\) is the cube of \(x\) and \(8\) is the cube of \(2\). Thus,

• \(x^3 + 8 = (x + 2)(x^2 - 2x + 4)\)

This factorization breaks the problem into more manageable parts. Recognizing a sum of cubes can simplify partial fraction decomposition and other algebraic processes.

System of Equations

A system of equations is a set of multiple equations that share common variables. Solving such systems means finding values of the variables that satisfy all equations simultaneously.
In the context of partial fraction decomposition, once the polynomial is broken down, you need to find constants (A, B, C, etc.) by solving a system.
From the original exercise, after setting up the partial fraction form and clearing the denominator, we had to solve the system:

• \(A + B = -5\)
• \(2B - 2A + C = 18\)
• \(4A + 2C = -4\)

This system of equations, arising from equating coefficients, helps us determine the specific values of A, B, and C which result in the correct decomposition.
Solving these equations typically involves substitution or elimination methods.

Irreducible Quadratic Factors

Irreducible quadratic factors are important in partial fraction decomposition. These are quadratic expressions that cannot be factored further into real linear factors. For instance, \(x^2 - 2x + 4\) in the exercise is an irreducible quadratic factor.
When dealing with such factors in partial fraction decomposition, you need to account for them by using a linear numerator in your fraction. In our exercise, this translates to setting up the partial fraction as \(\frac{Bx+C}{x^2-2x+4}\).
The irreducibility indicates that no real number solutions satisfy \(x^2 - 2x + 4 = 0\), which means it can't be simplified further over the real numbers, and must be used as it is within the decomposition process. Understanding this helps manage and simplify complex rational expressions effectively.