Problem 21 - College Algebra

Question

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

Step-by-Step Solution

Answer

\(\frac{x}{(x-2)^2} = \frac{1}{x-2} + \frac{2}{(x-2)^2}\).

Step 1: Identify the Denominator's Structure

The denominator of the fraction \((x-2)^2\) indicates there is a repeating linear factor \((x-2)\).

Step 2: Set Up Partial Fraction Form

Since \((x-2)\) is a repeating linear factor, the partial fraction decomposition will have the form:\[ \frac{x}{(x-2)^2} = \frac{A}{x-2} + \frac{B}{(x-2)^2} \] where \(A\) and \(B\) are constants to be determined.

Step 3: Eliminate Denominators

Multiply through by \((x-2)^2\) to clear the denominators:\[ x = A(x-2) + B \]

Step 4: Expand and Collect Terms

Expand the right side of the equation:\[ x = Ax - 2A + B \]

Step 5: Equate Coefficients

Compare coefficients from both sides of the equation:- The coefficient of \(x\) on the right side, which is \(A\), must equal 1 on the left side.- The constant terms \(-2A + B \) on the right must equal 0.This gives us the system of equations:1. \( A = 1 \)2. \( -2A + B = 0 \)

Step 6: Solve for A and B

Substitute \(A = 1\) into the second equation:\[ -2(1) + B = 0 \]\[ -2 + B = 0 \]\[ B = 2 \]Thus, \(A = 1\) and \(B = 2\).

Step 7: Write the Final Partial Fraction Decomposition

Substitute the values of \(A\) and \(B\) back into the partial fraction form:\[ \frac{x}{(x-2)^2} = \frac{1}{x-2} + \frac{2}{(x-2)^2} \]

Key Concepts

Repeating Linear Factors

When dealing with polynomial fractions, recognizing repeating linear factors is crucial. A repeating linear factor is present in the denominator of a fraction, and occurs when a linear factor appears more than once.
In the given exercise, the denominator is \((x-2)^2\), indicating the linear factor \((x-2)\) repeats. When the factor repeats, you expand the partial fraction to include multiple terms where each term represents a power of the repeated factor.

• First term: \(\frac{A}{x-2}\)
• Second term: \(\frac{B}{(x-2)^2}\)

This setup helps in determining coefficients easily—a key step in simplifying the fraction through partial fraction decomposition.

Coefficient Comparison

Once you've expanded the equation into a partial fraction, the next step is comparing coefficients. This process helps in identifying unknown constants in the expanded terms.
For example, if you multiply through by the original denominator to eliminate it, you'll end up with an equation containing different powers of \(x\). Here's how that plays out:

• From the equation \(x = Ax - 2A + B\), identify coefficients of like terms.
• The coefficient of \(x\) on the left is \(1\), so \(A\) must also equal \(1\) on the right.
• For the constant terms, compare \(-2A + B\) on the right side to the consistent term (\(0\) on the left).

By setting these equal, you can solve for unknown constants such as \(A\) and \(B\) involved in the original fraction decomposition.

Equation Solving

After identifying the coefficients, solve both equations simultaneously to find the unknowns. In this exercise, you are solving for \(A\) and \(B\). Based on the coefficient comparisons, you'll generally get a straightforward system of equations.

• Start by observing we have:
  • \(A = 1\)
  • \(-2A + B = 0\)
• Substitute: since \(A = 1\), insert this value into \(-2A + B = 0\) to find \(B\).
• This substitution leads to solving the simple linear equation: \(-2 + B = 0\).

From this equation, you find \(B = 2\). Solving these simple equations is the linchpin to successful partial fraction decomposition.

Denominator Structure

Understanding the structure of your denominator is foundational when working with partial fraction decomposition. Recognizing what type of factors you’re dealing with determines the form of the partial fraction.
With a denominator like \((x-2)^2\), you're dealing with a specific type of structure—the repeating linear factor. This structure dictates that your partial fraction decomposition needs to include separate terms for each instance of the factor.

• The simplest factor corresponds to \(\frac{1}{x-2}\).
• The repeating factor component is addressed by \(\frac{1}{(x-2)^2}\).

Grasping this structural insight smooths the process of writing out the terms needed for the corresponding partial fraction, ultimately making it easier to resolve unknown coefficients through comparison and equation solving.