Problem 19 - College Algebra with Corequisite Support

Question

For the following exercises, find the decomposition of the partial fraction for the nonrepeating linear factors. $$ \frac{3 x-1}{x^{2}-5 x+6} $$

Step-by-Step Solution

Answer

\( \frac{-5}{x-2} + \frac{8}{x-3} \) is the partial fraction decomposition.

Step 1: Factor the Denominator

First, you need to factor the denominator, which is a quadratic expression: \( x^2 - 5x + 6 \). To factor it, look for two numbers that multiply to \( 6 \) and add to \( -5 \). These numbers are \( -2 \) and \( -3 \). Thus, the factorization of the denominator is \( (x - 2)(x - 3) \).

Step 2: Set Up the Partial Fraction Decomposition

Since the denominator is factored as \((x - 2)(x - 3)\), the partial fraction decomposition for the rational expression \( \frac{3x - 1}{(x - 2)(x - 3)} \) will be: \( \frac{A}{x - 2} + \frac{B}{x - 3} \), where \( A \) and \( B \) are constants to be determined.

Step 3: Clear the Denominator

Multiply both sides of the equation by \((x - 2)(x - 3)\) to eliminate the denominators. You will get: \[ 3x - 1 = A(x - 3) + B(x - 2) \].

Step 4: Distribute and Combine Terms

Expand the terms on the right side: \( A(x - 3) = Ax - 3A \) and \( B(x - 2) = Bx - 2B \). This gives: \[ 3x - 1 = Ax - 3A + Bx - 2B \]. Combine like terms: \[ 3x - 1 = (A + B)x - (3A + 2B) \].

Step 5: Form Equations Based on Coefficients

Since the expression is equal to \( 3x - 1 \), match the coefficients of \( x \) and the constant term: \[ A + B = 3 \] and \[ -3A - 2B = -1 \].

Step 6: Solve the System of Equations

Solve the equations: \( A + B = 3 \) and \( -3A - 2B = -1 \). Substitute \( A = 3 - B \) into the second equation: \( -3(3 - B) - 2B = -1 \). Simplify to find \( B \): \[ -9 + 3B - 2B = -1 \] which simplifies to \[ B = 8 \]. Substitute back to find \( A \): \( A = 3 - 8 = -5 \).

Step 7: Write the Final Decomposition

Insert the values of \( A \) and \( B \) into the partial fraction decomposition: \( \frac{3x - 1}{x^2 - 5x + 6} = \frac{-5}{x - 2} + \frac{8}{x - 3} \).

Key Concepts

Quadratic Expression Factoring

When dealing with a quadratic expression such as \( x^2 - 5x + 6 \), factoring is a key step in simplifying it. First, identify two numbers that multiply to the last term, 6, and add up to the middle term's coefficient, -5. These numbers are -2 and -3, leading to the factorization \((x - 2)(x - 3)\).
Factoring helps break down the polynomial into simpler linear terms, which are easier to work with during partial fraction decomposition.
This technique is critical for converting complex fractions into simpler ones that contain nonrepeating linear factors.

Nonrepeating Linear Factors

A nonrepeating linear factor arises when each factor appears only once in the denominator. The expression \( \frac{3x - 1}{x^2 - 5x + 6} \) simplifies to \( \frac{3x - 1}{(x - 2)(x - 3)} \) after factoring.

The partial fraction decomposition for this results in \( \frac{A}{x - 2} + \frac{B}{x - 3} \). Here, \( A \) and \( B \) are constants. This setup is possible only because \((x-2)\) and \((x-3)\) are nonrepeating.

• Each linear factor has a distinct partial fraction associated with it.
• Decomposition into nonrepeating linear factors aids in simplification by allowing for easier integration and solution of equations.

System of Equations

Solving for the constants \( A \) and \( B \) in partial fraction decomposition leads to a system of equations. Once the denominators are cleared, you equate the coefficients from both sides of the equation.

Given the expression \( 3x - 1 = A(x - 3) + B(x - 2) \), we distribute and combine like terms:

• \( 3x - 1 = (A + B)x - (3A + 2B) \)

By matching the coefficients for \( x \) and the constant term:

• \( A + B = 3 \)
• \(-3A - 2B = -1\)

This forms a system of equations that, when solved, helps determine the values of \( A \) and \( B \). Solving these equations involves substitution and simplification, revealing \( A = -5 \) and \( B = 8 \).
Understanding how to solve this kind of system is crucial in conducting a partial fraction decomposition successfully.