Problem 6 - Calculus - AP Edition

Question

Write out the form of the partial fraction decomposition. (Do not find the numerical values of the coefficients.) $$\frac{3 x}{(x-1)\left(x^{2}+6\right)}$$

Step-by-Step Solution

Answer

The decomposition is \(\frac{A}{x-1} + \frac{Bx+C}{x^2+6}\).

Step 1: Analyze the Denominator

The given fraction has the denominator \((x-1)(x^2 + 6)\). It consists of two distinct factors: a linear factor \((x-1)\) and a quadratic factor \((x^2 + 6)\).

Step 2: Set Up Partial Fractions

For partial fraction decomposition, each distinct factor in the denominator corresponds to a term in the decomposition. A linear factor \((x-1)\) corresponds to a term \(\frac{A}{x-1}\), while the irreducible quadratic factor \((x^2 + 6)\) corresponds to a term \(\frac{Bx + C}{x^2 + 6}\).

Step 3: Write the Decomposition

Combine the terms from Step 2 to form the partial fraction decomposition: \[ \frac{3x}{(x-1)(x^2+6)} = \frac{A}{x-1} + \frac{Bx+C}{x^2+6} \] Here, \(A\), \(B\), and \(C\) are the coefficients that need to be determined.

Key Concepts

Linear Factor

In partial fraction decomposition, identifying linear factors is essential to setting up the decomposition properly. A linear factor is a polynomial factor of degree one. It generally looks like \(ax + b\), where \(a\) and \(b\) are constants.
In the exercise example, the linear factor is \(x - 1\). This factor is straightforward, involving only \(x\) to the power of one.

When setting up partial fractions, each linear factor in the denominator corresponds to a simple term of the form \(\frac{A}{x-1}\). Here, \(A\) represents a constant that will be determined later.

Key points about linear factors in decomposition:

• They contribute simple terms to the decomposition.
• These terms are relatively easy to handle since they're linear.
• Finding the coefficient involves basic algebraic manipulation.

Understanding linear factors is your first step in breaking down complex fractions into simpler parts.

Quadratic Factor

A quadratic factor is a polynomial factor of degree two, typically written as \(ax^2 + bx + c\). Such factors add another layer of complexity to partial fraction decomposition.
In our exercise, the quadratic factor is \(x^2 + 6\), which can't be simplified further because it has no real roots.

When dealing with quadratic factors, each corresponds to a term of the form \(\frac{Bx + C}{x^2 + 6}\) in the decomposition. Here, \(B\) and \(C\) are constants that are yet to be found.

Some important notes about quadratic factors:

• They complicate the decomposition since they involve two coefficients, \(B\) and \(C\).
• Being irreducible means you’ll work with complex algebra or calculus to solve for the coefficients.
• These factors often require techniques such as completing the square or comparing coefficients.

Excel in handling quadratic factors to strengthen your understanding of fraction decomposition.

Denominator Analysis

Before diving into partial fraction decomposition, it’s crucial to conduct a thorough denominator analysis. This process involves understanding the structure of the denominator and identifying each distinct factor.

In the given exercise, the denominator \((x-1)(x^2+6)\) consists of a linear factor \((x-1)\) and a quadratic factor \((x^2+6)\). Breaking these down:

• The linear factor \((x-1)\) suggests a simple term in the decomposition.
• The quadratic factor \((x^2+6)\), with no real roots, requires a more complex term.

Denominator analysis guides the structure of your partial fraction setup.

Key steps for effective denominator analysis include:

• Identify and label each distinct factor in the denominator.
• Classify whether factors are linear or quadratic based on their degree.
• Recognize irreducible quadratics to anticipate the need for solving more complex equations.

By starting with a thorough denominator analysis, you set a solid foundation for successful partial fraction decomposition.