Problem 54 - College Algebra

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 the Denominator Factors

The first step is to recognize the factors in the denominator. Here, the denominator is \((x^2 + 2x)^2\). This is an irreducible quadratic term raised to the power of 2, suggesting a repeating irreducible quadratic factor.

Step 2: Set Up the Partial Fraction Decomposition

For the decomposition, since the denominator is \(((x^2 + 2x)^2)\), which consists of a repeated irreducible quadratic factor, the partial fraction decomposition format is:\[\frac{5x^3 - 2x + 1}{(x^2 + 2x)^2} = \frac{Ax + B}{x^2 + 2x} + \frac{Cx + D}{(x^2 + 2x)^2}\]

Step 3: Clear the Denominator

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

Step 4: Expand the Terms

Expand the terms on the right side:\[(Ax + B)(x^2 + 2x) = Ax^3 + 2Ax^2 + Bx^2 + 2Bx\]Thus, the equation becomes:\[5x^3 - 2x + 1 = Ax^3 + (2A + B)x^2 + 2Bx + Cx + D\]

Step 5: Collect Like Terms

Combine the like terms on the right side:\[5x^3 - 2x + 1 = Ax^3 + (2A + B)x^2 + (2B + C)x + D\]

Step 6: Solve for Coefficients

To find the coefficients, equate the corresponding coefficients from both sides:1. \( A = 5 \)2. \( 2A + B = 0 \)3. \( 2B + C = -2 \)4. \( D = 1 \)Now, solve the equations:- From 1: \( A = 5 \)- Substitute \( A = 5 \) into 2: \( 10 + B = 0 \Rightarrow B = -10 \)- Substitute \( B = -10 \) into 3: \(-20 + C = -2 \Rightarrow C = 18 \)- From 4: \( D = 1 \)

Step 7: Write the Final Decomposition

Substitute the values of the coefficients into the partial fraction decomposition:\[\frac{5x^3 - 2x + 1}{(x^2 + 2x)^2} = \frac{5x - 10}{x^2 + 2x} + \frac{18x + 1}{(x^2 + 2x)^2}\]

Key Concepts

Irreducible Quadratic Factor

An irreducible quadratic factor is a quadratic expression that cannot be factored into the real-number field. To determine if a quadratic is irreducible, you must try to factor it. If it resists all attempts, and its discriminant (given by \( b^2 - 4ac \) in the quadratic formula) is negative, it remains in its quadratic form. In the given exercise, the expression \((x^2 + 2x)^2\) contains the irreducible quadratic \(x^2 + 2x\). Since the discriminant \(b^2 - 4ac\) equals \(4 - 4 \times 1 \times 0 = 4\), it does not suggest irreducibility at first glance since the discriminant is zero. Nonetheless, for partial fraction decomposition, this factor is treated as irreducible because it does not separate into linear parts within the context here, particularly since we focus on systematic decomposition for educational instances.

Repeated Root Quadratic

A repeated root quadratic factor occurs when a quadratic is raised to a power higher than one. This means that the quadratic appears more than once in the factored form of a polynomial. In our exercise, the factor \( (x^2 + 2x) ^ 2 \) is a quadratic repeated twice. The fact that it is squared means we'll encounter multiple terms in our decomposition — specifically one term for the quadratic level and another for the squared version. This characterization is vital for setting up the decomposition structure, enabling the breakdown of complex rational expressions into simpler fractions.

Algebraic Fractions Decomposition

Algebraic fraction decomposition involves breaking a complex fraction into simpler parts. This is particularly useful for integral calculations or revealing simpler behavior in expressions.In the given exercise, we set up the decomposition for the expression \( \frac{5x^3 - 2x + 1}{(x^2 + 2x)^2} \). Due to the repeated irreducible quadratic factor, our setup includes:

• Numerator \( Ax + B \) for the single power part, \( x^2 + 2x \)
• Numerator \( Cx + D \) for the squared part, \( (x^2 + 2x)^2 \)

The target is to express the polynomial as a sum of simpler fractions, where each component fits the denominator's degree appropriate to its individual factor power.

Solving Coefficients in Partial Fractions

Solving for coefficients in partial fraction decomposition is critical to unlocking the simplified expression. This is done by equating corresponding coefficients from the expanded equation.For our task, we equate:

• \( A \) with the coefficient of \( x^3 \) term, resulting in \( A = 5 \)
• Twice \( A \) plus \( B \) for \( x^2 \), giving \( 2 \times 5 + B = 0 \Rightarrow B = -10 \)
• Twice \( B \) plus \( C \) for \( x \), leading to \( 2(-10) + C = -2 \Rightarrow C = 18 \)
• \( D \) equated directly to the constant term, thus \( D = 1 \)

Substituting these values back furnishes the entire decomposition. This approach requires meticulous alignment of power terms, ensuring precision in final formulations.