Problem 47 - College Algebra with Corequisite Support

Question

For the following exercises, find the decomposition of the partial fraction for the irreducible repeating quadratic factor. $$ \frac{x^{2}+5 x+5}{(x+2)^{2}} $$

Step-by-Step Solution

Answer

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

Step 1: Identify the Type of Partial Fraction Decomposition

Given the function \( \frac{x^{2}+5 x+5}{(x+2)^{2}} \), notice that the denominator \((x+2)^{2}\) is a repeated linear factor. To perform a partial fraction decomposition, we express the rational expression as a sum of fractions where the denominators are powers of these linear factors.

Step 2: Set Up Partial Fraction Decomposition

The given rational function can be decomposed into partial fractions as follows:\[\frac{x^{2}+5x+5}{(x+2)^{2}} = \frac{A}{x+2} + \frac{B}{(x+2)^{2}}\]where \(A\) and \(B\) are constants to be determined.

Step 3: Clear the Denominator

Multiply through the equation by the denominator \((x+2)^{2}\) to eliminate the fractions:\[x^{2} + 5x + 5 = A(x+2) + B\]

Step 4: Expand and Combine Terms

Expand the terms on the right side:\[A(x+2) + B = Ax + 2A + B\]This simplifies to: \[Ax + 2A + B\]

Step 5: Equate and Solve for Coefficients

Now, compare the coefficients of like terms from both sides of the expanded equation:1. For the \(x\) term: \( A = 1 \)2. For the constant term: \(2A + B = 5\)Using these equations, substitute \(A = 1\) into the second equation:\[2(1) + B = 5\]Thus, \(B = 3\).

Step 6: Write Resulting Partial Fractions

Substitute \(A = 1\) and \(B = 3\) back into the partial fraction decomposition, yielding:\[\frac{x^{2}+5x+5}{(x+2)^{2}} = \frac{1}{x+2} + \frac{3}{(x+2)^{2}}\]

Key Concepts

Repeated Linear Factor

In the world of partial fraction decomposition, a repeated linear factor plays a crucial role. It refers to a factor in the denominator of a rational function that appears more than once. For example, in the expression \((x+2)^2\), \(x+2\) is our linear factor and is repeated, hence the name "repeated linear factor."

When decomposing a rational expression containing such factors, we write a series of fractions. Each fraction has as its denominator one of the powers of the repeated linear factor. This approach ensures that all potential simplifications are considered in the decomposition process.

• For \((x+2)^2\), our decomposed form would have denominators of \(x+2\) and \((x+2)^2\).
• This makes it possible to use coefficients for each fraction to fully represent the original rational expression.

Rational Expression

A rational expression is simply a fraction with polynomials in both the numerator and denominator. In the example given, \(\frac{x^2 + 5x + 5}{(x+2)^2}\), we see that the numerator \(x^2 + 5x + 5\) is a polynomial of degree two, while the denominator \((x+2)^2\) is the square of the linear polynomial \(x+2\).

Rational expressions lend themselves well to a mathematical technique known as partial fraction decomposition. This technique helps us break down a complicated fraction into simpler parts, which are easier to work with, especially when integrating or calculating limits. Partial fractions are particularly useful in calculus, where they can transform a complex integral into more manageable pieces.

• The key to decomposing the rational expression is to express it in terms of simpler fractions whose denominators derive from the factors of the original denominator.
• This task often involves separating the expression into distinct pieces, simplifying both integration and differentiation processes further along.

Expanding and Simplifying

Once you've set up your partial fraction decomposition, the next steps are to expand and simplify the expression. In our example, we express the decomposition as \(\frac{A}{x+2} + \frac{B}{(x+2)^2}\).

To clear the fractions, we multiply both sides by the common denominator, here \((x+2)^2\). Doing this yields an equation free of fractions: \(x^2 + 5x + 5 = A(x+2) + B\).

The expansion step involves distributing \(A\) through \(x+2\), giving \(Ax + 2A + B\).

• This turns the equation into a polynomial identity, equating coefficients with those on the left side \(x^2 + 5x + 5\).
• Expanding correctly is crucial to accurately match the coefficients on both sides and solve for unknown variables \(A\) and \(B\).

Solving for Coefficients

In the final step of partial fraction decomposition, we solve for the coefficients. This involves comparing the expanded and simplified expression with the original to determine the values of the unknown coefficients \(A\) and \(B\).

Inspect the equation \(Ax + 2A + B = x^2 + 5x + 5\) and equate the coefficients of like terms:

• For the \(x\) terms, compare the coefficients directly: \(A = 1\).
• For the constant terms, set the equation: \(2A + B = 5\).

Substitute \(A = 1\) into the constant equation:

• \(2(1) + B = 5\)
• Solve to get \(B = 3\).

These steps now allow us to compose the simplified form: \(\frac{1}{x+2} + \frac{3}{(x+2)^2}\). Solving for coefficients gives precise tuning needed to align the partial fractions with the original expression.