Problem 16 - College Algebra

Question

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

Step-by-Step Solution

Answer

\(\frac{2x - 3}{x^2 - 6x + 5} = \frac{1}{4(x-1)} + \frac{7}{4(x-5)}\)

Step 1: Factor the Denominator

Let's start by factoring the quadratic denominator. The given denominator is \(x^2 - 6x + 5\). To factor this, we need two numbers that multiply to \(5\) and add up to \(-6\). The numbers \(-1\) and \(-5\) satisfy this, so \(x^2 - 6x + 5 = (x-1)(x-5)\).

Step 2: Setup the Partial Fraction Decomposition

Since the denominator factors as \( (x-1)(x-5) \), the partial fraction decomposition can be set up as: \[ \frac{2x - 3}{(x-1)(x-5)} = \frac{A}{x-1} + \frac{B}{x-5} \]where \(A\) and \(B\) are constants we need to determine.

Step 3: Clear the Fractions

Multiply both sides by the common denominator \((x-1)(x-5)\) to clear the fractions, resulting in:\[ 2x - 3 = A(x-5) + B(x-1)\]

Step 4: Expand and Combine Like Terms

Expand the right-hand side of the equation:\[ A(x-5) + B(x-1) = Ax - 5A + Bx - B \]Combine like terms:\[ 2x - 3 = (A + B)x - (5A + B) \]

Step 5: Match Coefficients

To find \(A\) and \(B\), match the coefficients of corresponding terms on both sides. For the terms in \(x\): \( A + B = 2 \).For the constant terms:\(-5A - B = -3 \).

Step 6: Solve the System of Equations

We have the system of equations: 1. \( A + B = 2 \) 2. \(-5A - B = -3 \)Solve the first equation for \(B\):\[ B = 2 - A \] Substitute \(B = 2 - A\) into the second equation:\[-5A - (2 - A) = -3\]Simplify:\[-5A - 2 + A = -3\]Combine like terms:\[-4A - 2 = -3\]Solve for \(A\):\[-4A = -1\]\[A = \frac{1}{4}\]Substitute back to find \(B\):\[B = 2 - \frac{1}{4} = \frac{7}{4}\]

Step 7: Write the Final Decomposition

Substitute \(A\) and \(B\) back into the partial fraction decomposition:\[ \frac{2x - 3}{x^2 - 6x + 5} = \frac{\frac{1}{4}}{x-1} + \frac{\frac{7}{4}}{x-5} \]This can also be written as:\[ \frac{2x - 3}{x^2 - 6x + 5} = \frac{1}{4(x-1)} + \frac{7}{4(x-5)} \]

Key Concepts

Nonrepeating Linear Factors

In partial fraction decomposition, nonrepeating linear factors refer to factors in the denominator that occur exactly once and involve a variable. They are essential to breaking down complex fractions into simpler, more manageable parts. When the denominator of a rational function can be factored into distinct linear factors, each factor gets its own partial fraction in the decomposition.

For example, consider the function \( \frac{2x - 3}{(x-1)(x-5)} \). Here, \(x-1\) and \(x-5\) are nonrepeating linear factors because each appears once in the factored form of the denominator. This simplifies the task of decomposition, as it means we place each factor in the denominator of a separate fraction. We then assign coefficients (such as \(A\) and \(B\)) to these fractions and solve for these constants.

The equation would then look like \( \frac{A}{x-1} + \frac{B}{x-5} \). Understanding how to identify and handle nonrepeating linear factors is crucial for mastering partial fraction decomposition efficiently.

System of Equations

A system of equations is a set of two or more equations with the same variables. They are used in partial fraction decomposition to determine the coefficients of the decomposed fractions. Once you break the fractions apart, you equate the original expression to the sum of the fractions. Then by matching the coefficients on both sides of the equation, you generate a system of linear equations to solve for the unknowns, commonly represented by letters like \(A\) and \(B\).

In the example of \( \frac{2x - 3}{(x-1)(x-5)} \), the relation \(2x - 3 = A(x-5) + B(x-1)\) unfolds into a system of equations when matching coefficients:

• \(A + B = 2\) for the \(x\) terms
• \(-5A - B = -3\) for the constant terms.

Solving these equations will give the values of \(A\) and \(B\). Being adept at setting up and solving systems of equations ensures you can efficiently find the coefficients needed for accurate decomposition.

Factoring Quadratics

Factoring quadratics is a method used to break down a quadratic expression into two simpler binomials. The skill of factoring is pivotal in the process of partial fraction decomposition, especially when dealing with quadratic denominators since partial fraction decomposition requires the factorization of the denominator.

In our example, the original denominator \(x^2 - 6x + 5\) needs to be factored to work with the expression. The general approach involves finding two numbers that multiply to the constant term and add up to the linear coefficient. For \(x^2 - 6x + 5\), these numbers are \(-1\) and \(-5\), hence we factor it as \((x - 1)(x - 5)\).

Successfully factoring the quadratic is the gateway to rewriting the fraction into partial fractions. Mastery of this concept ensures you can tackle a wide range of algebraic problems that involve polynomial expressions as denominators.