Problem 22 - College Algebra

Question

For the following exercises, find the decomposition of the partial fraction for the repeating linear factors. $$\frac{7 x+14}{(x+3)^{2}}$$

Step-by-Step Solution

Answer

The partial fraction decomposition is \(\frac{7}{x+3} - \frac{7}{(x+3)^2}\).

Step 1: Understand the Problem

We need to decompose the fraction \(\frac{7x+14}{(x+3)^2}\) into partial fractions. The denominator \((x+3)^2\) is a repeated linear factor.

Step 2: Set Up the Partial Fraction Decomposition

For a repeated linear factor, the partial fraction decomposition should take the form: \(\frac{A}{x+3} + \frac{B}{(x+3)^2}\). We need to find the constants \(A\) and \(B\).

Step 3: Combine the Right Side to a Single Fraction

Write the partial fractions as a single fraction: \(\frac{A}{x+3} + \frac{B}{(x+3)^2} = \frac{A(x+3) + B}{(x+3)^2}\). This results in \(\frac{Ax + 3A + B}{(x+3)^2}\).

Step 4: Equate and Solve for Coefficients

Equate \(\frac{Ax + 3A + B}{(x+3)^2}\) to \(\frac{7x + 14}{(x+3)^2}\). From this, you get: \(Ax + 3A + B = 7x + 14\).

Step 5: Form Two Equations to Solve for A and B

By comparing coefficients, equate: \(A = 7\) (coefficient of \(x\)) and \(3A + B = 14\) (constant term).

Step 6: Solve the Equations

Use \(A = 7\) in the equation \(3A + B = 14\) to find \(B\). Substitute: \(3 \times 7 + B = 14\), thus \(21 + B = 14\). Hence, \(B = 14 - 21 = -7\).

Step 7: Write the Final Decomposition

The partial fraction decomposition is \(\frac{7}{x+3} + \frac{-7}{(x+3)^2}\). This can be simplified to \(\frac{7}{x+3} - \frac{7}{(x+3)^2}\).

Key Concepts

Repeated Linear Factors

When working with partial fraction decomposition, understanding repeated linear factors is crucial. In our example, the expression \((x+3)^2\) is a repeated linear factor. This means that the factor \(x+3\) appears more than once in the denominator, specifically squared in this case.
In partial fractions, the general rule for dealing with repeated linear factors \((x+a)^n\) involves writing each power of the factor up to \(n\) with separate fractions. So if we have \((x+3)^2\), the decomposition needs a term for both \(x+3\) and \((x+3)^2\).

• Separate terms: \(\frac{A}{x+3}\) and \(\frac{B}{(x+3)^2}\).
• Each coefficient (like \(A\) and \(B\)) is determined by solving a system of equations derived from equating the given fraction to its decomposed form.

Repeated linear factors add complexity to fraction decomposition but follow a formalized method to handle them effectively.

Constant Coefficients

In the process of fraction decomposition, finding constant coefficients is a key step. These coefficients represent the numerators in the decomposed fractional expression.
Returning to our example, we are solving for coefficients \(A\) and \(B\). These constants align with particular terms in your original fraction, ensuring both sides are equivalent throughout the equation.

• For example: \(A\) corresponds with the linear part of the numerator.
• \(B\) represents the additional constant term needed to fully match the original numerator upon expansion.

By solving a set of equations derived from the decomposition, you can determine these values. In this problem, solving resulted in \(A=7\) and \(B=-7\). These solutions helped reformulate the original function into its decomposed form.

Fraction Decomposition

Fraction decomposition essentially breaks down a complex fraction into simpler components. This is particularly useful in integration and algebra for simplifying problems.
The main goal in fraction decomposition is to transform an expression into the sum or difference of simpler fractions with constant coefficients in the numerators.

• Identify underlying factors in the denominator.
• Delineate those factors into individual fractions, termed partial fractions.
• Calculate constants to equate both sides of the expression.

In practical terms, think of fraction decomposition as a puzzle. Each piece — the partial fractions and their coefficients — comes together to recreate the original picture, or fraction.

Algebraic Equations

Algebraic equations are essential for finding the unknown constants during fraction decomposition.
When decomposing fractions, we create an algebraic equation by setting up an equality between the rational expression and its intended decomposed form. From here, the equivalence of coefficients method helps us compare similar terms on both sides of the equation.

• Include all terms in the comparison: corresponding powers of \(x\), and constant terms.
• This results in a system of linear equations.
• Solution of these equations gives the necessary coefficients for the decomposition.

For the given problem, comparing coefficients after standardizing the expression yielded two equations: \(A = 7\) and \(3A + B = 14\). Solving these equations via substitution or elimination revealed the values of \(A\) and \(B\), completing the decomposition task.