Problem 22 - College Algebra

Question

Find the decomposition of the partial fraction for the repeating linear factors. \(\frac{7 x+14}{(x+3)^{2}}\)

Step-by-Step Solution

Answer

\(\frac{7}{x+3} - \frac{7}{(x+3)^2}\)

Step 1: Identify the Denominator

The denominator of the expression is \((x+3)^2\). This tells us that the repeated linear factor is \(x+3\). We also note that it repeats twice.

Step 2: Setup the Partial Fraction

Since we have a repeating linear factor \((x+3)\) repeated twice, our partial fraction decomposition will be of the form: \[\frac{A}{x+3} + \frac{B}{(x+3)^2}\]where \(A\) and \(B\) are constants we need to determine.

Step 3: Write the Equation

We equate the original fraction to our partial fraction form:\[\frac{7x + 14}{(x+3)^2} = \frac{A}{x+3} + \frac{B}{(x+3)^2}\]

Step 4: Eliminate the Denominator

Multiply both sides by \((x+3)^2\) to eliminate the denominator:\[7x + 14 = A(x+3) + B\]

Step 5: Expand and Simplify

Expand the equation: \[7x + 14 = Ax +3A + B\]

Step 6: Equate Coefficients

To solve for \(A\) and \(B\), equate the coefficients of \(x\) and the constant terms:1. For \(x\) terms: \(7 = A\)2. Constant terms: \(14 = 3A + B\)

Step 7: Solve for Constants

Substitute \(A = 7\) into the constant terms equation:\[14 = 3(7) + B\]Thus, \(B = 14 - 21 = -7\).

Step 8: Write the Decomposition

The partial fraction decomposition is:\[\frac{A}{x+3} + \frac{B}{(x+3)^2} = \frac{7}{x+3} - \frac{7}{(x+3)^2}\]

Key Concepts

Repeating Linear Factors

In partial fraction decomposition, identifying repeating linear factors is crucial. They occur when we have factors in the denominator that appear more than once. For instance, in the expression \((x+3)^2\), the factor \(x+3\) repeats. This indicates a special way to set up our decomposition.
For every repeated linear factor \((x+a)^n\), you must write a series of terms in your decomposition:

• \( \frac{A_1}{x + a} \)
• \( \frac{A_2}{(x + a)^2} \)
• Continue this pattern up to \( \frac{A_n}{(x + a)^n} \)

Each term has a separate numerator, indicating constants that need solving. In our case, we write \( \frac{A}{x+3} + \frac{B}{(x+3)^2} \) as the partial fraction form because our factor \(x+3\) repeats twice.

Coefficient Comparison

After setting up the partial fractions, we need to determine the values of the constants involved, like \(A\) and \(B\). This is where the method of coefficient comparison comes into play.
Once the denominators are cleared by multiplying both sides by the common denominator, we compare the coefficients of corresponding terms on both sides of the equation. For example, from the equation:\[7x + 14 = Ax + 3A + B\]

• Equate coefficients for \(x\): Since we have \(7x\) on the left, we set \(A = 7\).
• Equate constant terms: From \(14\) on the left side, equate to \(3A + B\) and solve.

The coefficients are like puzzle pieces, helping us piece together the values that fill in our partial fractions accurately.

Algebraic Fractions

Algebraic fractions are the backbone of partial fraction decomposition. Breaking down complex fractions into simpler, manageable parts allows more straightforward integration or further algebraic manipulation.
An algebraic fraction is simply a fraction that contains a polynomial in its numerator, denominator, or both. In partial fractions, we look at how to express such fractions as a sum of simpler fractions. For instance, in our task, the initial fraction \(\frac{7x + 14}{(x+3)^2}\) is split into \(\frac{A}{x+3} + \frac{B}{(x+3)^2}\). This transformation simplifies calculations, especially in calculus contexts where these expressions often appear.
Mastering decomposition equips students to handle quadratic or even higher degree fractions efficiently, providing a toolkit for tackling a wide range of mathematical problems.

Identifying Denominators

Identifying and analyzing the denominator is the first crucial step in partial fraction decomposition. Recognizing whether factors are linear, quadratic, or have multiplicity impacts how we set up the decomposition.
The original exercise requires us to inspect \( (x+3)^2 \). Notice:

• The factor \(x+3\) is linear because it's to the first power.
• The exponent \(2\) shows we deal with repeated factors.

Understanding the structure of the denominator dictates the necessary form for partial fractions. This structured approach helps maintain accuracy when solving for unknown constants and ensures all bases are covered in diverse algebraic scenarios.