Problem 19 - College Algebra

Question

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

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

Step 1: Factor the Denominator

The given expression is \( \frac{3x-1}{x^2 - 5x + 6} \). First, we need to factor the denominator, \( x^2 - 5x + 6 \). Finding factors of 6 that add up to -5, we get -2 and -3. So, the factorization is \((x-2)(x-3)\).

Step 2: Set Up Partial Fraction Decomposition

Since the factors are linear and distinct, the partial fraction decomposition of \( \frac{3x-1}{(x-2)(x-3)} \) is \( \frac{A}{x-2} + \frac{B}{x-3} \). We need to find constants \( A \) and \( B \).

Step 3: Write the Equation for the Decomposition

Combine the partial fractions into a single fraction: \( \frac{A(x-3) + B(x-2)}{(x-2)(x-3)} = \frac{3x-1}{(x-2)(x-3)} \). This implies that \( A(x-3) + B(x-2) = 3x-1 \).

Step 4: Solve for Constants A and B

Expand and equate coefficients: \( Ax - 3A + Bx - 2B = 3x - 1 \). Comparing coefficients: \( A + B = 3 \) (for \( x \)) and \( -3A - 2B = -1 \) (constant term). Solve this system of equations.

Step 5: Solve the System of Equations

Using the equations \( A + B = 3 \) and \( -3A - 2B = -1 \), we can substitute \( B = 3 - A \) into the second equation: \(-3A - 2(3 - A) = -1\). Simplify to get \( -3A - 6 + 2A = -1 \), leading to \( -A = 5 \) so \( A = -5 \). Substitute back to find \( B = 8 \).

Step 6: Write the Final Decomposition

Substitute \( A \) and \( B \) back into the partial fractions. The decomposition is \( \frac{-5}{x-2} + \frac{8}{x-3} \).

Key Concepts

Linear Factors

Linear factors are expressions of the form \(x - r\) where \(r\) is a constant. In the context of polynomial expressions, identifying linear factors is vital. They represent the roots or solutions of the polynomial equation when set to zero.

For example, if you consider the quadratic polynomial \(x^2 - 5x + 6\), it's factored into linear factors as \(x - 2\) and \(x - 3\). This process shows that the solutions to the polynomial \(x^2 - 5x + 6 = 0\) are \(x = 2\) and \(x = 3\). Recognizing these linear factors allows us to decompose rational expressions efficiently into simpler parts in partial fraction decomposition.

Factoring Polynomials

Factoring polynomials involves expressing a polynomial as a product of its factors, which ideally should be polynomials of lower degrees. The goal is to break down the expression into simpler parts that can be easily managed.

For example, the expression \(x^2 - 5x + 6\) is a quadratic polynomial. We factor it by looking for two numbers that multiply to 6 (the constant term) and add up to -5 (the coefficient of the middle term). Here, -2 and -3 fit the requirements. Therefore, \(x^2 - 5x + 6\) can be rewritten as \( (x - 2)(x - 3)\).

• This process simplifies complex expressions into simpler linear factors.
• It allows for easier computation, especially in integration and solving equations.

System of Equations

A system of equations is a collection of two or more equations with the same set of variables. In the case of partial fraction decomposition, a system of equations is used to find the constants in the numerators of the decomposed fractions.

For the example given in the problem, after setting the partial fractions, we derived, \(A(x-3) + B(x-2) = 3x - 1\), leading to a system of two equations:

• \(A + B = 3\)
• \(-3A - 2B = -1\)

By solving this system, we determine the values of \(A\) and \(B\), which in this particular case are \(-5\) and \(8\) respectively. Solving systems of equations involves methods such as substitution, elimination, or matrix operations, depending on complexity.

Rational Expressions

Rational expressions are fractions where both the numerator and the denominator are polynomials. Understanding them is essential because they often appear in calculus and algebra problems.

In the context of partial fraction decomposition, you take a complex rational expression and break it down into simpler fractions, making the whole expression easier to integrate or differentiate.

• The expression \(rac{3x-1}{x^2 - 5x + 6}\) is a rational expression.
• By decomposing it into \(rac{-5}{x-2} + rac{8}{x-3}\), it becomes easier to handle.

A rational expression is deemed simple when its numerator has a lower degree than its denominator. Techniques like partial fraction decomposition are crucial in simplifying and integrating these expressions in advanced mathematics.