Problem 10 - College Algebra

Question

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

\(\frac{1}{2x + 5} - \frac{1}{3x + 5}\)

Step 1: Factor the Denominator

To decompose the fraction, first factor the denominator. The expression is \(6x^2 + 25x + 25\). We look for factors of the form \((ax + b)(cx + d)\). Using the quadratic formula or factorization methods, the factors of our quadratic are \((2x + 5)(3x + 5)\).

Step 2: Set Up Partial Fractions

Now that the denominator is factored, express the fraction as a sum of partial fractions: \[ \frac{x}{6x^2 + 25x + 25} = \frac{A}{2x + 5} + \frac{B}{3x + 5} \] where \(A\) and \(B\) are constants to be determined.

Step 3: Clear the Denominator

Multiply through by the common denominator \((2x + 5)(3x + 5)\) to eliminate the denominators: \[ x = A(3x + 5) + B(2x + 5) \].

Step 4: Expand and Combine Like Terms

Expand both terms on the right-hand side: \[ x = (3A)x + 5A + (2B)x + 5B \]. Combine like terms: \[ x = (3A + 2B)x + (5A + 5B) \].

Step 5: Equate Coefficients

For the equation to hold for all \(x\), the coefficients of corresponding powers of \(x\) on both sides must be equal: - Coefficient of \(x\): \(3A + 2B = 1\)- Constant term: \(5A + 5B = 0\).

Step 6: Solve for Variables A and B

From the equations:1. \(3A + 2B = 1\)2. \(5A + 5B = 0\) (simplified: \(A + B = 0\))Solve the second equation for \(A\): \[ A = -B \]. Substitute \(A = -B\) into the first equation:\[ 3(-B) + 2B = 1 \rightarrow -3B + 2B = 1 \rightarrow -B = 1 \rightarrow B = -1 \]. Substitute back to find \(A\): \[ A = -(-1) = 1 \].

Step 7: Write the Final Partial Fraction Decomposition

Using the values of \(A\) and \(B\), the partial fraction decomposition of the expression is: \[ \frac{x}{6x^2 + 25x + 25} = \frac{1}{2x + 5} - \frac{1}{3x + 5} \].

Key Concepts

Quadratic Factorization

When dealing with partial fractions, the first step often involves quadratic factorization. This means breaking down a quadratic expression into two factors that can be multiplied together to yield the original quadratic. For example, when we have the polynomial \(6x^2 + 25x + 25\), we need to find two binomials whose product gives us this polynomial. This involves identifying numbers that satisfy both the sum and the product requirements determined by the original quadratic's coefficients. Using techniques like the quadratic formula or trial and error, we eventually discover that the factors are \((2x + 5)(3x + 5)\). This vital step simplifies the overall fraction and enables us to use the method of partial fraction decomposition.

Linear Factors

Linear factors are expressions like \(ax + b\), which represent straight lines when graphed. In the context of partial fraction decomposition, once we have factored a quadratic polynomial into linear factors, each factor forms a part of the decomposed fractions. For the given exercise, the quadratic in the denominator \(6x^2 + 25x + 25\) was factored into the linear factors \((2x + 5)\) and \((3x + 5)\). Each of these factors becomes the denominator of a separate fraction, allowing us to write our expression as a sum of simpler fractions: \(\frac{A}{2x + 5} + \frac{B}{3x + 5}\). This step is crucial as it enables easier handling of the fraction when solving for unknown coefficients.

Coefficient Comparison

Coefficient comparison is a method used to solve for unknowns in an equation that results from decomposing a fraction. After expressing the fraction in terms of its linear factors and multiplying through to eliminate the denominators, we get an equation in \(x\), like \(x = (3A + 2B)x + (5A + 5B)\). To find the values of \(A\) and \(B\), the coefficients of similar terms on both sides of the equation must be equal. This gives us a system of equations, which is solved to find \(A = 1\) and \(B = -1\). This method leverages the idea that for an equation to be true for all values of \(x\), the coefficients of corresponding powers of \(x\) on both sides must match.

Algebraic Decomposition

Algebraic decomposition involves breaking down more complex expressions into simpler ones. For partial fractions, this means taking a single rational expression—like the one in the problem \(\frac{x}{6x^2 + 25x + 25}\)—and expressing it as the sum of simpler fractions with linear denominators.

• Begin by factoring the denominator.
• Express the fraction in terms of constants \(A\) and \(B\).
• Clear the fractions by eliminating the denominators.
• Use coefficient comparison to solve for \(A\) and \(B\).

Each of these fractions is easier to integrate, differentiate, or otherwise manipulate. The final decomposition \(\frac{1}{2x + 5} - \frac{1}{3x + 5}\) makes many calculus operations, specifically integration, much more straightforward, thus highlighting the importance of this technique.