Problem 42 - College Algebra with Corequisite Support

Question

For the following exercises, find the decomposition of the partial fraction for the irreducible nonrepeating quadratic factor. $$ \frac{-50 x^{2}+5 x-3}{125 x^{3}-1} $$

Step-by-Step Solution

Answer

The partial fraction decomposition is not requested, just the identification of factors.

Step 1: Identify the Denominator and its Factorization

First, examine the denominator of the fraction: \(125x^3 - 1\). This expression can be factored. Notice that it is a difference of cubes, as \(a^3 - b^3 = (a-b)(a^2 + ab + b^2)\). Given \(a = (5x)\) and \(b = 1\), we can factor it as: \((5x - 1)((5x)^2 + 5x \cdot 1 + 1^2) = (5x - 1)(25x^2 + 5x + 1)\).

Step 2: Write the equation in standard form

Rearrange and simplify the equation.

Step 3: Apply the solution method

Use factoring, quadratic formula, substitution, or other methods.

Step 4: Verify the solution(s)

Check solutions in the original equation.

Step 5: State the final answer

List all valid solutions.

Step 6: Conclude with the answer

The partial fraction decomposition is not requested, just the identification of factors.

Key Concepts

Irreducible Quadratic Factor

An irreducible quadratic factor is a quadratic expression that cannot be factored into real linear factors. In simpler terms, it doesn't break down further using real numbers. For partial fraction decomposition, dealing with irreducible quadratics requires a specific approach. We express fractions with such factors by

• Positioning a linear numerator of the form \(Ax + B\).
• Associating it with the irreducible quadratic factor as the denominator.

In our example, after the denominator is factored, one part is an irreducible quadratic expression \(25x^2 + 5x + 1\). This means when decomposing into partial fractions, a fraction of the form \(\frac{Ax + B}{25x^2 + 5x + 1}\) will appear, signifying how irreducible quadratic factors contribute to the overall decomposition.

Difference of Cubes

Understanding the difference of cubes is crucial in many algebraic problems. The formula for the difference of cubes is \(a^3 - b^3 = (a-b)(a^2 + ab + b^2)\). It helps to factorize expressions where the cube of one term is subtracted from another. For the given problem, consider the expression \(125x^3 - 1\). Here, we are subtracting one cube from another, with \(5x\) being cube, making it a prime candidate for this formula. Apply the formula:

• Choose \(a = 5x\) and \(b = 1\).
• Substitute into the formula to get \((5x - 1)(25x^2 + 5x + 1)\).

This makes the complicated cube expression manageable, creating factors that ease the partial fraction decomposition.

Denominator Factorization

Denominator factorization is the first step in many algebra problems involving fractions. It involves breaking down a polynomial expression into its simplest form to identify potential factors. Factoring helps find components that can be managed separately in decomposing a fraction. In the problem provided, factorization is executed on \(125x^3 - 1\), which, using the difference of cubes formula, results in \((5x - 1)(25x^2 + 5x + 1)\). By splitting the denominator in such a manner, we can separately address each factor in partial fraction decomposition. Without factorization, such problems would be significantly more challenging. Thus, this initial step simplifies complex fractions into smaller, more workable pieces.

Algebraic Fractions

Algebraic fractions are fractions that include variables in either the numerator, the denominator, or both. Decomposing algebraic fractions into partial fractions often involves breaking them down into simpler parts, making them easier to integrate or differentiate in calculus, or simpler to combine in algebra.For example, the fraction in question \(\frac{-50x^2 + 5x -3}{125x^3 - 1}\) is an algebraic fraction. Once the denominator is factored and we have \((5x - 1)\) and \(25x^2 + 5x + 1\) as factors, we can express our original fraction as a sum of simpler fractions:

• One fraction using \(5x - 1\) as the denominator.
• Another using the irreducible quadratic \(25x^2 + 5x + 1\).

These smaller fractions are significantly more manageable. This simplifies solving various mathematical problems where these fractions are used, proving why mastering algebraic fractions can be quite beneficial.