Problem 17 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 1-28: Find the partial fraction decomposition. $$ \frac{x^{2}+x-6}{\left(x^{2}+1\right)(x-1)} $$

Step-by-Step Solution

Answer

The partial fraction decomposition is \( \frac{3x + 4}{x^2 + 1} - \frac{2}{x - 1} \).

Step 1: Determine the Form of Partial Fractions

Given the function \( \frac{x^2 + x - 6}{(x^2 + 1)(x-1)} \), we note that the denominator can be factored into two parts: \( x^2 + 1 \) and \( x - 1 \). Since \( x^2 + 1 \) is an irreducible quadratic factor, we assign partial fractions as: \( \frac{Ax + B}{x^2 + 1} + \frac{C}{x - 1} \).

Step 2: Set Up the Equation

We rewrite the function as \[ \frac{x^2 + x - 6}{(x^2 + 1)(x-1)} = \frac{Ax + B}{x^2 + 1} + \frac{C}{x - 1} \]. Then, multiply through by the common denominator \((x^2 + 1)(x - 1)\) to obtain: \[ x^2 + x - 6 = (Ax + B)(x - 1) + C(x^2 + 1) \].

Step 3: Expand and Simplify

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

Step 4: Compare Coefficients

By comparing coefficients from both sides, we have: \( 1 = A + C \), \( 1 = B - A \), and \(-6 = -B + C \).

Step 5: Solve the System of Equations

Solve the system of equations:1. \( A + C = 1 \)2. \( B - A = 1 \)3. \( -B + C = -6 \).From equation 2, \( B = A + 1 \).Substitute \( B = A + 1 \) into equation 3: \(-A - 1 + C = -6 \) leads to \( -A + C = -5 \).Now solve \(\begin{cases} A + C = 1 \cr -A + C = -5 \end{cases}\).Add these equations: \(2C = -4\) which gives \(C = -2\).Substitute \(C = -2\) into \(A + C = 1\): \(A - 2 = 1\), hence \(A = 3\).Use \(B = A + 1\) to find \(B = 4\).

Step 6: Write the Partial Fraction Decomposition

Substitute \( A = 3 \), \( B = 4 \), and \( C = -2 \) back into the original form: \[ \frac{x^2 + x - 6}{(x^2 + 1)(x-1)} = \frac{3x + 4}{x^2 + 1} - \frac{2}{x - 1} \].

Key Concepts

Irreducible Quadratic Factor

When working with partial fraction decomposition, one encounters various types of factors in the denominator. Among these is the irreducible quadratic factor. So, what makes a quadratic factor irreducible? It's a polynomial of degree two that cannot be factored further into real linear factors. In simpler terms, it means there's no way to break it down into simpler real number parts further.

An example of this from the exercise is the factor \(x^2 + 1\). Unlike \(x^2 - 1\), which can be factored into \((x + 1)(x - 1)\), \(x^2 + 1\) remains "whole". Crucially, the irreducible term enters the partial fraction as a ratio of a linear numerator over the quadratic denominator; hence we see \(\frac{Ax + B}{x^2 + 1}\).

Partial fraction decomposition requires an understanding of such fractions, as ignoring their unique nature can lead to incorrect results. Remember:

• The linear term in the numerator reflects the degree of the quadratic term in the denominator.
• Always check if a quadratic is irreducible by attempting to factor it over real numbers.
• The roots of irreducible quadratics, if they exist, are complex numbers.

Comparing Coefficients

The technique of comparing coefficients is essential in determining unknowns in partial fraction decomposition. Once you have multiplied through by the common denominator and expanded the equation, the next step is to line up terms by their degrees and compare these to terms on the other side.

In the exercise, after expansion, we got: \((A+C)x^2 + (B-A)x + (-B+C)\). Each of these terms has to match its counterpart from the original expression \(x^2 + x - 6\). By setting the coefficients equal, a system of linear equations is established:

• The coefficients of \(x^2\): \(A + C = 1\)
• The coefficients of \(x\): \(B - A = 1\)
• The constant terms: \(-B + C = -6\)

This process allows us to create a blueprint for solving the unknown variables \(A\), \(B\), and \(C\), ensuring the two sides of the equation reflect the same thing.

If you're ever confused in this step, remember:

• The left-hand side and right-hand side of your equation need to be identical in terms of coefficients.
• Each equation gives a valuable clue that will guide you to the correct values of your unknowns.
• Double-check each term to avoid assigning incorrect coefficients.

System of Equations

In partial fraction decomposition, pulling together and solving a system of equations can appear daunting, but let's break it down. Once you've compared coefficients and generated equations, you now have to solve for the unknowns. This exercise provides these equations:

• \(A + C = 1\)
• \(B - A = 1\)
• \(-B + C = -6\)

At first glance, these can seem complex, but they're your path to finding \(A\), \(B\), and \(C\).

Start by isolating one variable in one equation and substituting these back into the others. For instance, \(B\) is made expressible through \(A\) in equation \(B = A + 1\). This makes it easier to substitute and solve further, allowing us to solve for \(C\) and eventually \(A\).

With systematic substitution and arriving at simpler equations, solving becomes straightforward. Remember:

• A structured approach to solving can prevent confusion.
• Each substitution should be cross-referenced with the equations to ensure consistency.
• Patience in detailing your steps helps uncover any mishaps before they compound.