Problem 41 - College Algebra with Corequisite Support

Question

For the following exercises, find the decomposition of the partial fraction for the irreducible nonrepeating quadratic factor. $$ \frac{4 x^{2}+4 x+12}{8 x^{3}-27} $$

Step-by-Step Solution

Answer

The decomposition is \( \frac{1}{2x-3} + \frac{1}{4x^2+6x+9} \).

Step 1: Factor the Denominator

The denominator is given as \( 8x^3 - 27 \). Notice that this is a difference of cubes, which can be factored using the formula, \( a^3 - b^3 = (a-b)(a^2 + ab + b^2) \). Here, \( a = 2x \) and \( b = 3 \). Therefore, \( 8x^3 - 27 = (2x - 3)(4x^2 + 6x + 9) \).

Step 2: Set Up Partial Fraction Decomposition

Given the factored denominator \((2x - 3)(4x^2 + 6x + 9)\), the fraction \( \frac{4x^2 + 4x + 12}{(2x - 3)(4x^2 + 6x + 9)} \) can be decomposed into partial fractions as: \( \frac{A}{2x-3} + \frac{Bx + C}{4x^2 + 6x + 9} \).

Step 3: Express the Equation

Multiply through by the denominator \((2x-3)(4x^2+6x+9)\) to eliminate the fractions. This results in: \[ 4x^2 + 4x + 12 = A(4x^2+6x+9) + (Bx+C)(2x-3). \]

Step 4: Expand and Collect Like Terms

Expand both sides: \( A(4x^2+6x+9) = 4Ax^2 + 6Ax + 9A \) and \( (Bx+C)(2x-3) = 2Bx^2 + (2C-3B)x - 3C \). Combine these to yield: \( 4x^2 + 4x + 12 = (4A + 2B)x^2 + (6A + 2C - 3B)x + (9A - 3C) \).

Step 5: Solve for Coefficients

Equate coefficients: \( 4A + 2B = 4 \), \( 6A + 2C - 3B = 4 \), \( 9A - 3C = 12 \). Solving these, we find \( A = 1, B = 0 \), and \( C = 1 \).

Step 6: Write Final Partial Fraction Decomposition

Substitute back the values obtained for A, B, and C into the partial fraction decomposition: \( \frac{1}{2x-3} + \frac{0 \cdot x + 1}{4x^2 + 6x + 9} \), simplifying this gives: \( \frac{1}{2x-3} + \frac{1}{4x^2 + 6x + 9} \).

Key Concepts

Difference of Cubes

Understanding the process of factoring a difference of cubes can simplify complex polynomial expressions. A difference of cubes is expressed in the form \( a^3 - b^3 \), and its factorization is given by the formula:

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

In this formula, the factor \( (a-b) \) is linear, while \( (a^2 + ab + b^2) \) is a quadratic expression. Consider the denominator in the exercise: \( 8x^3 - 27 \). Identifying \( a = 2x \) and \( b = 3 \), we factor it as:

• \( (2x - 3)(4x^2 + 6x + 9) \)

Use this method to simplify fractions and make them easier to manipulate in algebraic operations.

Irreducible Quadratic Factor

An irreducible quadratic factor is a second-degree polynomial that cannot be further factored over the real numbers. In our exercise, after factoring the difference of cubes, we encounter the quadratic \( 4x^2 + 6x + 9 \). To determine if a quadratic is irreducible, perform a check using the discriminant \( b^2 - 4ac \). If the result is negative, the quadratic does not factor further:

• For \( 4x^2 + 6x + 9 \), compute the discriminant: \( 6^2 - 4 \times 4 \times 9 = 36 - 144 = -108 \)
• Since it is negative, \( 4x^2 + 6x + 9 \) is irreducible over the reals.

Knowing whether a quadratic is irreducible aids in setting up partial fraction decomposition correctly and helps simplify complex algebraic fractions.

Coefficient Matching

Coefficient matching is a crucial technique used in partial fraction decomposition to find unknown constants. Once the expression is expanded, match coefficients of corresponding powers of \( x \) on both sides of the equation. This involves writing each side of the equation with similar terms:

• Given \( 4x^2 + 4x + 12 = (4A + 2B)x^2 + (6A + 2C - 3B)x + (9A - 3C) \)

Align terms based on powers of \( x \), forming a system:

• \( 4A + 2B = 4 \)
• \( 6A + 2C - 3B = 4 \)
• \( 9A - 3C = 12 \)

Solving these equations gives the values of \( A, B, C \). Coefficient matching, therefore, is essential for identifying specific values in the decomposition process.

Algebraic Fractions

Algebraic fractions involve expressions where the numerator and denominator are polynomials. To decompose these fractions, especially when dealing with irreducible factors, utilize partial fraction decomposition. This breaks down a complex fraction into a sum of simpler ones, easing integration or differentiation tasks.In our exercise, the fraction \( \frac{4x^2 + 4x + 12}{(2x-3)(4x^2 + 6x + 9)} \) splits into simpler fractions:

• \( \frac{A}{2x-3} \) and \( \frac{Bx + C}{4x^2 + 6x + 9} \)

This enables easier manipulation and can simplify solving equations, making understanding algebraic relationships straightforward. Mastering these concepts helps in tackling broader algebraic problems effectively.