Problem 10 - College Algebra

Question

For the following exercises, find the decomposition of the partial fraction for the nonrepeating linear factors. $$\frac{x}{6 x^{2}+25 x+25}$$

Step-by-Step Solution

Answer

\( \frac{-1}{3x+5} + \frac{1}{2x+5} \) is the decomposition.

Step 1: Identify the Denominator

First, we need to factor the denominator of the fraction. The given denominator is \(6x^2 + 25x + 25\). We will begin by identifying if this can be factored into linear factors for partial fraction decomposition.

Step 2: Factor the Denominator into Linear Factors

The expression \(6x^2 + 25x + 25\) can be factored as \((3x + 5)(2x + 5)\). This was determined by trial and error with values that multiply to \(6 \times 25\) and sum to \(25\). These serve as the non-repeating linear factors of the denominator.

Step 3: Write the Partial Fraction Decomposition

Set up the partial fraction decomposition. Since the denominator has factors \((3x + 5)\) and \((2x + 5)\), the decomposition takes the form: \(\frac{x}{(3x+5)(2x+5)} = \frac{A}{3x+5} + \frac{B}{2x+5}\). Here, \(A\) and \(B\) are constants we need to determine.

Step 4: Clear the Denominator and Combine Terms

Multiply through by the common denominator \((3x+5)(2x+5)\) to eliminate fractions: \[x = A(2x+5) + B(3x+5)\]. Expand the right-hand side: \[x = (2A + 3B)x + (5A + 5B)\].

Step 5: Equate Coefficients and Solve for Constants

From \(x = (2A + 3B)x + (5A + 5B)\), equate the coefficients of like terms:- Coefficient of \(x\): \(1 = 2A + 3B\)- Constant term: \(0 = 5A + 5B\)From the second equation, \( A + B = 0 \). Substitute \(A = -B\) into the first equation to solve: \[ 1 = 2(-B) + 3B \] \[ 1 = -2B + 3B \] \[ B = 1 \]. Then, \(A = -1\) since \(A = -B\).

Step 6: Write the Final Decomposed Form

With \(A = -1\) and \(B = 1\), the partial fraction decomposition is: \[ \frac{x}{(3x+5)(2x+5)} = \frac{-1}{3x+5} + \frac{1}{2x+5} \].

Key Concepts

Nonrepeating Linear Factors

Nonrepeating linear factors are an essential part of partial fraction decomposition. They refer to factors of a polynomial that are linear and do not repeat. In mathematical terms, a linear factor is an expression of the form \(ax + b\), where \(a\) and \(b\) are constants and \(a\) is not zero. For effective decomposition, each factor must be distinct. This ensures each part of the fraction corresponds to a unique solution.

• In our exercise, the factors \((3x + 5)\) and \((2x + 5)\) are linear because they fall into the structure \(ax + b\).
• Also, they are nonrepeating as each factor appears only once in the denominator.

Understanding nonrepeating linear factors is crucial because they dictate the setup of the partial fraction decomposition. This setup is the basis for finding the constants that will complete the solution.

Denominator Factoring

Denominator factoring is the process of breaking down a polynomial into a product of its linear and/or irreducible quadratic factors. This step is vital in partial fraction decomposition to simplify the fraction and identify its individual components.

• In our given problem, the denominator \(6x^2 + 25x + 25\) needed to be factored first.
• Through a mix of analytic techniques and trial methods, it was factored into \((3x + 5)(2x + 5)\).

Factoring the denominator correctly is essential. It enables the decomposition into simpler parts, making the complex fraction more manageable. This method is useful in a variety of mathematical applications, including solving differential equations and evaluating integrals.

Equating Coefficients

Equating coefficients is a method used to find unknown constants in algebraic equations. Once a polynomial equation is expanded, coefficients of like terms on both sides of the equation are set equal to each other. This is because if two polynomials are equal, then the coefficients of their corresponding terms must also be equal.

• In the partial fraction decomposition, the original equation was expanded: \(x = (2A + 3B)x + (5A + 5B)\).
• The coefficients of \(x\) terms were equated: \(1 = 2A + 3B\).
• The constant terms were set equal: \(0 = 5A + 5B\).

This method allows us to set up a system of linear equations. Solving these equations, we discover the numerical values for the constants \(A\) and \(B\). It provides clarity and a systematic approach to finding these constants, which are integral to the correct partial fraction decomposition.

Constants Determination

Determining constants is the final step in the process of partial fraction decomposition. These constants are crucial as they finalize the expression and ensure accuracy in solutions. After equating the coefficients, we are left with a system of linear equations.

• From the second equation, we had \(A + B = 0\), leading to \(A = -B\).
• Substituting this into the first equation: \(1 = -2B + 3B\). This upon simplification gives \(B = 1\).
• Substituting \(B = 1\) back gives \(A = -1\).

Thus, the constants \(A\) and \(B\) were found to be \(-1\) and \(1\), respectively. These constants allow the final decomposition of the fraction accurately. The process of determining constants is a systematic approach that guarantees precision and aids in further applications of the decomposed expression.