Problem 54 - College Algebra with Corequisite Support

Question

For the following exercises, find the decomposition of the partial fraction for the irreducible repeating quadratic factor. $$ \frac{5 x^{3}-2 x+1}{\left(x^{2}+2 x\right)^{2}} $$

Step-by-Step Solution

Answer

The decomposition is \( \frac{5x - 10}{x^2 + 2x} + \frac{18x + 1}{(x^2 + 2x)^2} \).

Step 1: Identify Denominator Factors

The denominator of the given fraction is \( (x^2 + 2x)^2 \). Here, \( x^2 + 2x \) is an irreducible quadratic factor that repeats.

Step 2: Set Up Partial Fraction Decomposition

For an irreducible repeating quadratic factor \( (x^2 + 2x) \), the form is: \[\frac{5x^3 - 2x + 1}{(x^2 + 2x)^2} = \frac{Ax + B}{x^2 + 2x} + \frac{Cx + D}{(x^2 + 2x)^2}\] where \(A, B, C,\) and \(D\) are constants to be determined.

Step 3: Clear the Denominator

Multiply both sides by \( (x^2 + 2x)^2 \) to clear the fraction: \[5x^3 - 2x + 1 = (Ax + B)(x^2 + 2x) + (Cx + D)\]

Step 4: Expand and Combine Like Terms

Expand the terms and collect like terms on the right side: \[(Ax + B)(x^2 + 2x) = Ax^3 + 2Ax^2 + Bx^2 + 2Bx\]So, combine: \[= Ax^3 + (2A + B)x^2 + 2Bx\]

Step 5: Compare Coefficients

By comparing coefficients from both sides of the equation,\(Ax^3 + (2A + B)x^2 + 2Bx + Cx + D = 5x^3 - 2x + 1\)Equating coefficients gives us:1. \( A = 5 \)2. \( 2A + B = 0 \)3. \( 2B + C = -2 \)4. \( D = 1 \)

Step 6: Solve the System of Equations

Using the system of equations established:1. \( A = 5 \)2. \( 2(5) + B = 0 \rightarrow B = -10 \)3. \( 2(-10) + C = -2 \rightarrow C = 18 \)4. \( D = 1 \)Plug these values back into the partial fraction form.

Step 7: Final Partial Fraction Decomposition

Substituting values, the partial fraction decomposition is:\[\frac{5x^3 - 2x + 1}{(x^2 + 2x)^2} = \frac{5x - 10}{x^2 + 2x} + \frac{18x + 1}{(x^2 + 2x)^2}\]

Key Concepts

Irreducible Quadratic Factor

In partial fraction decomposition, an irreducible quadratic factor is a quadratic expression that cannot be factored into real linear factors. This means the expression doesn’t break down into simpler multiplicative components involving real numbers. For example, in our given fraction, the denominator involves the term \(x^2 + 2x\). At a glance, it seems it could be factored given its structure, but it turns out to be irreducible over the real numbers in its current form.

When you encounter such quadratics, you'll have to handle them with special care during decomposition. The formula for partial fractions involving an irreducible quadratic looks like: \(\frac{Ax + B}{x^2 + 2x}\), where \(A\) and \(B\) are constants.

This form is necessary because it keeps the decomposition manageable and correctly aligned with the quadratic's degree, ensuring no further simplification is accidentally omitted.

Repeating Factors

Repeating factors in the denominator imply that the factor occurs more than once. In our exercise, we have the factor \((x^2 + 2x)\) which is squared, indicating it repeats. Recognizing when a factor repeats is crucial because it dictates the structure of the partial fraction decomposition.

When a factor repeats, your decomposition must include forms for each repetition up to its power. For example, since \((x^2 + 2x)^2\) is a second-degree repeat, the setup involves both \(\frac{Ax + B}{x^2 + 2x}\) and \(\frac{Cx + D}{(x^2 + 2x)^2}\) to account for each instance of the quadratic.

This approach ensures every power of the repeating factor is represented, allowing the entire expression to be accurately reconstructed using the determined constants.

System of Equations

A system of equations emerges naturally during partial fraction decomposition as we equate coefficients. Once you clear the denominators and combine like terms on both sides, you end up with a polynomial on each side of the equation.

Our example generates the following equations from comparing coefficients:

• \(A = 5\)
• \(2A + B = 0\)
• \(2B + C = -2\)
• \(D = 1\)

Solving this system involves substituting known values step-by-step to find the unknown coefficients.

Begin with the simplest equations and move to the more complex ones, substituting already found values to reduce complexity and avoid errors.

Coefficient Comparison

Coefficient comparison is a vital tool in partial fraction decomposition, enabling us to find the constants. Once the decomposition is set up and cleared of denominators, expand both sides and equate terms of equal powers.

For instance, in \(Ax^3 + (2A + B)x^2 + 2Bx + Cx + D = 5x^3 - 2x + 1\), we compare each coefficient aligned with their corresponding powers of \(x\). This comparison gives individual equations:

• \(A = 5\)
• \(2A + B = 0\)
• \(2B + C = -2\)
• \(D = 1\)

The straightforward nature of comparison makes it a powerful method to determine unknowns, delineating concrete pathways to balance expressions effectively.