Problem 49 - College Algebra

Question

Find the decomposition of the partial fraction for the irreducible repeating quadratic factor. \(\frac{x^{2}+25}{\left(x^{2}+3 x+25\right)^{2}}\)

Step-by-Step Solution

Answer

The partial fraction decomposition is \(\frac{x}{x^2 + 3x + 25} - \frac{3x}{(x^2 + 3x + 25)^2}\).

Step 1: Identifying the Denominator Type

First, we observe that the denominator \((x^2 + 3x + 25)^2\) is a repeated irreducible quadratic factor. Therefore, we follow the partial fraction decomposition for repeated irreducible quadratic factors.

Step 2: Setting Up the Partial Fraction Decomposition

The structure for the partial fraction decomposition of a repeating irreducible quadratic \((x^2 + 3x + 25)^2\) is \[\frac{Ax + B}{x^2 + 3x + 25} + \frac{Cx + D}{(x^2 + 3x + 25)^2}.\]Here, \(A, B, C,\) and \(D\) are constants that we need to determine.

Step 3: Clearing the Denominator

Multiply both sides by \((x^2 + 3x + 25)^2\) to eliminate the denominators, resulting in: \[x^2 + 25 = (Ax + B)(x^2 + 3x + 25) + (Cx + D).\]

Step 4: Expanding the Right Side

Expand \((Ax + B)(x^2 + 3x + 25)\): \[(Ax + B)(x^2 + 3x + 25) = Ax^3 + (3A + B)x^2 + (25A + 3B)x + 25B.\]Substitute this back into the equation to get: \[x^2 + 25 = Ax^3 + (3A + B)x^2 + (25A + 3B)x + (25B + Cx + D).\]

Step 5: Matching Coefficients

Now match coefficients of \(x^3, x^2, x,\) and the constant term from both sides:1. Coefficient of \(x^3\): \(A = 0\).2. Coefficient of \(x^2\): \(3A + B = 1\), with \(A = 0\), hence \(B = 1\).3. Coefficient of \(x\): \(25A + 3B + C = 0\), with \(A = 0\) and \(B = 1\), hence \(C = -3\).4. Constant term: \(25B + D = 25\), with \(B = 1\), hence \(D = 0\).

Step 6: Writing the Final Decomposition

Substitute the calculated values \(A = 0, B = 1, C = -3, D = 0\) into the partial fraction form:\[\frac{x}{x^2 + 3x + 25} - \frac{3x}{(x^2 + 3x + 25)^2}.\]This represents the partial fraction decomposition of the given expression.

Key Concepts

Irreducible Quadratic Factor

In algebra, an irreducible quadratic factor is a quadratic expression that cannot be factored further into real numbers. This means that the expression

• has a degree of 2, and
• does not simplify into simpler terms using real coefficients.

An example of an irreducible quadratic is \(x^2 + 3x + 25\). It doesn't break down into simpler linear factors. It might seem complicated when you face it within a problem, but recognizing these factors helps set the stage for solving algebraic equations with complex denominators. In partial fraction decomposition, spotting irreducible quadratics accurately is crucial for structuring the initial decomposition attempts correctly.

Repeated Factors

Repeated factors in a denominator occur when a particular factor appears more than once. In such cases, as seen in the term \((x^2 + 3x + 25)^2\), the denominator consists of the same quadratic factor repeated. This repetition requires a specific technique in partial fraction decomposition.
The presence of repeated factors changes the way we express the fractions. For example, rather than simply having one term for the quadratic, we develop additional terms corresponding to the number of times it repeats. This is evident in our setup:

• \(\frac{Ax + B}{x^2 + 3x + 25}\)
• \(\frac{Cx + D}{(x^2 + 3x + 25)^2}\)

Each fraction corresponds to a power of the repeating factor, allowing us to handle the complexity effectively by correctly aligning terms and clearly organizing the decomposition process.

Coefficient Matching

Coefficient matching is a method used to determine the unknown constants in a partial fraction decomposition. When you've cleared denominators and have set up your equations, you'll obtain a polynomial equation which involves matching coefficients from both sides.
Here, the equation derived from clearing and expanding is:\[x^2 + 25 = Ax^3 + (3A + B)x^2 + (25A + 3B)x + (25B + Cx + D)\]We equate the coefficients for the same powers of \(x\) on both sides:

• \(x^3: A = 0\)
• \(x^2: 3A + B = 1\) leading to \(B = 1\)
• \(x: 25A + 3B + C = 0\) resulting in \(C = -3\)
• Constant term: \(25B + D = 25\) where \(D = 0\)

Through this careful matching, you systematically find the values for \(A, B, C,\) and \(D\) that satisfy the equation.

Algebraic Fractions

Algebraic fractions are fractions where the numerator and the denominator are algebraic expressions. These expressions may include any combination of constants and variables. When we're solving problems involving algebraic fractions, especially in partial fraction decomposition, it's necessary to apply algebraic rules strategically.
The goal is often to decompose complex fractions into simpler ones. By doing this, each can be easier to integrate or simplify further. Using properties of the denominator, such as factorizable elements, greatly aids in the deconstruction process. In our case, by breaking \(\frac{x^2 + 25}{(x^2 + 3x + 25)^2}\) into simpler terms like

• \(\frac{x}{x^2 + 3x + 25}\)
• \(- \frac{3x}{(x^2 + 3x + 25)^2}\)

we enable streamlined calculations or integrals involving the expression, thereby revealing clearer insights into the mathematical operations.