Problem 37 - College Algebra

Question

For the following exercises, find the decomposition of the partial fraction for the irreducible non repeating quadratic factor. $$\frac{4 x^{2}+5 x+3}{x^{3}-1}$$

Step-by-Step Solution

Answer

The partial fraction decomposition is \( \frac{4}{x - 1} + \frac{1}{x^2 + x + 1} \).

Step 1: Recognize the Denominator

First, recognize that the denominator can be factored using the identity for the difference of cubes: \[ x^3 - 1 = (x - 1)(x^2 + x + 1) \]So, we will decompose the fraction \( \frac{4x^2 + 5x + 3}{(x - 1)(x^2 + x + 1)} \).

Step 2: Set Up the Partial Fraction Decomposition

The decomposition of the fraction will be of the form:\[ \frac{4x^2 + 5x + 3}{(x - 1)(x^2 + x + 1)} = \frac{A}{x - 1} + \frac{Bx + C}{x^2 + x + 1} \]Here, \( A \), \( B \), and \( C \) are constants that we need to determine.

Step 3: Clear the Fractions

Multiply both sides by the common denominator \((x - 1)(x^2 + x + 1)\) to clear the fractions:\[ 4x^2 + 5x + 3 = A(x^2 + x + 1) + (Bx + C)(x - 1) \]

Step 4: Expand and Collect Like Terms

Expand the right side:\[ A(x^2 + x + 1) = Ax^2 + Ax + A \]\[ (Bx + C)(x - 1) = Bx^2 - Bx + Cx - C \]Combine these:\[ Ax^2 + Ax + A + Bx^2 - Bx + Cx - C \]Collect like terms:\[ (A + B)x^2 + (A - B + C)x + (A - C) \]

Step 5: Set Up and Solve the System of Equations

Equate the coefficients from both sides:- For \(x^2\): \[ A + B = 4 \]- For \(x\): \[ A - B + C = 5 \]- For the constant term: \[ A - C = 3 \]Now, solve this system of equations to find \( A \), \( B \), and \( C \).

Step 6: Solve for A, B, and C

From the system, \( A + B = 4 \) and \( A - C = 3 \). Solving the third equation \( A - B + C = 5 \) as follows:1. Substitute from \( A - C = 3 \) gives \( C = A - 3 \).2. Substitute into \( A - B + C = 5 \), we get \( A - B + (A - 3) = 5 \), leading to \( 2A - B - 3 = 5 \). Therefore, \( 2A - B = 8 \).3. Using equations: - \( A + B = 4 \) - \( 2A - B = 8 \)Add both equations to get \( 3A = 12 \), so \( A = 4 \).4. Substitute \( A = 4 \) back into \( A + B = 4 \) to find \( B = 0 \).5. Finally, using \( A - C = 3 \), substituting \( A = 4 \) results in \( C = 1 \).

Step 7: Write the Final Decomposition

Substitute \( A = 4 \), \( B = 0 \), and \( C = 1 \) back into the partial fraction decomposition:\[ \frac{4x^2 + 5x + 3}{(x - 1)(x^2 + x + 1)} = \frac{4}{x - 1} + \frac{0x + 1}{x^2 + x + 1} \]This simplifies to:\[ \frac{4}{x - 1} + \frac{1}{x^2 + x + 1} \]

Key Concepts

Irreducible Quadratic Factors

When tackling partial fraction decomposition, recognizing irreducible quadratic factors in the denominator is key. An irreducible quadratic factor is a polynomial of degree 2 that cannot be factored further into linear factors with real coefficients. These appear when the quadratic equation does not have real roots. In our original problem, after factoring the denominator, we encounter the factor \(x^2 + x + 1\). Since this quadratic has no real roots, it remains as an irreducible quadratic factor.
This requires a specific form in our partial fraction decomposition, specifically \(\frac{Bx + C}{x^2 + x + 1}\), rather than just \(\frac{B}{x^2 + x + 1}\), because we cannot further break it down using real numbers. This characterizes how important it is to understand the role of irreducibility in setting up the decomposition correctly.

Difference of Cubes

The original problem's denominator is distilled using the difference of cubes formula. This formula helps in factorization when dealing with expressions like \(x^3 - 1\). Recognizing such a pattern is crucial because it allows us to simplify the expression and make the decomposition process more straightforward.

The difference of cubes formula is given by:

• \(x^3 - a^3 = (x - a)(x^2 + ax + a^2)\)

For our problem, it translates to \(x^3 - 1\) being \((x - 1)(x^2 + x + 1)\). This factorization highlights an essential step in resolving the denominator, serving as a foundation for building the partial fraction decomposition.

Algebraic Manipulation

In a nutshell, algebraic manipulation involves rearranging and simplifying algebraic expressions to achieve a desired form. For partial fraction decomposition, this initially includes clearing fractions by multiplying through by the common denominator.

In our example, we multiplied both sides of the equation by \((x-1)(x^2+x+1)\) to clear fractions. On the right-hand side, algebraic manipulation involves expanding products and then combining like terms.

• First, expand the expressions \(A(x^2 + x + 1)\) and \((Bx + C)(x - 1)\).
• After expanding, combine terms, for example, like\(Ax^2 + Bx^2\).

This step is essential because it leads us to create the system of equations needed to find the constants \(A\), \(B\), and \(C\). It emphasizes the practicality of orderly procedures in algebraic manipulation.

System of Equations

Solving the partial fraction decomposition requires us to set up a system of equations derived from matching coefficients. You equate the expanded and combined terms of the polynomial to the numerator parts like \(4x^2 + 5x + 3\) - thus ensuring agreement across the polynomial's degrees.

Our example results in:

• \(A + B = 4\) (for \(x^2\))
• \(A - B + C = 5\) (for \(x\))
• \(A - C = 3\) (constant terms)

Using these equations, solve step by step to determine \(A\), \(B\), and \(C\). This step by finding each unknown provides a concrete solution to our decomposition. Heading effectively into this phase shows the power of organizing and correlating information abstracted from algebra techniques to solve such systems.