Question

Resolve \(\frac{2 x+3}{x^{2}-6 x+5}\) into partial fractions. (1) \(\frac{1}{4}\left(\frac{13}{x-5}-\frac{5}{x-1}\right)\) (2) \(\frac{5}{x-5}-\frac{13}{4(x-1)}\) (3) \(\frac{13}{5(x-1)}-\frac{4}{5(x-5)}\) (4) \(\frac{5}{(x-5)}-\frac{4}{(x-1)}\)

Short answer

Question: Resolve the given fraction into partial fractions: \(\frac{2x+3}{x^2-6x+5}\) Answer: \(\frac{2x+3}{x^2-6x+5} =\frac{1}{4}\left(\frac{13}{x-5}-\frac{5}{x-1}\right)\)

Step-by-step solution

Factor the Denominator

Factor the denominator \(x^2-6x+5\): \((x^2-6x+5)=(x-1)(x-5)\).

Set up Partial Fractions

Set up the partial fractions with unknown constants A and B: \(\frac{2x+3}{(x-1)(x-5)} = \frac{A}{x-1} + \frac{B}{x-5}\).

Clear the Denominator

Clear the denominator by multiplying both sides of the equation by the factored denominator \((x-1)(x-5)\): \((2x+3)=A(x-5)+B(x-1)\).

Solve for Constants A and B

We can solve for A and B using substitution or a system of linear equations. In this case, we will use substitution. Let x = 1, then: \(A(1-5) + B(1-1) = 2(1)+3\) \(-4A = 5\) \(A = -\frac{5}{4}\) Let x = 5, then: \(A(5-1) + B(5-5) = 2(5)+3\) \(4A = 13\) \(A = \frac{13}{4}\)

Write the Final Answer

Substitute the values of A and B back into the partial fractions: \(\frac{2x+3}{(x-1)(x-5)} = \frac{-\frac{5}{4}}{x-1} + \frac{\frac{13}{4}}{x-5}\) The correct answer is (1): \(\frac{2x+3}{(x-1)(x-5)} =\frac{1}{4}\left(\frac{13}{x-5}-\frac{5}{x-1}\right)\).

Key concepts

Algebraic Fractions

Algebraic fractions are expressions that contain polynomials in their numerator, denominator, or both, similar to ordinary fractions that have numbers on the top and the bottom. When dealing with algebraic fractions, we often encounter complex-looking expressions that need simplification.

An example of an algebraic fraction is \( \frac{2x + 3}{x^2 - 6x + 5} \) from our exercise. The goal with such fractions is typically to simplify them or break them down into simpler parts. This process can involve partial fraction decomposition, which is a way to express the algebraic fraction as a sum of simpler fractions with easier-to-handle denominators and numerators. It's a useful technique in calculus and algebra as it can simplify integration and other operations.

During the learning process, it's crucial to be comfortable working with these algebraic structures since they form the basis of more advanced mathematics, including calculus. As a strategy to endure understanding, students should practice algebraic fraction simplification, look for common factors, and explore different methods of solving for variables present in the fractions.

Factorization

Factorization is a fundamental concept in algebra where you express a polynomial as the product of its factors. These factors are usually polynomials that are of lower degree than the original polynomial and are easier to work with. For example, factorization turns \( x^2 - 6x + 5 \) into \( (x - 1)(x - 5) \).

The purpose of factorization is to simplify complex expressions and solve equations more easily. It plays a crucial role in partial fraction decomposition, where we first factor the denominator before splitting the fraction into parts. This is crucial because the ability to see a polynomial as a product of its factors is the first step in breaking down algebraic fractions into partial fractions.

It's essential for students to learn and recognize common factorization patterns such as the difference of squares, trinomials, or sum and difference of cubes. Developing a strong foundation in factoring helps in not only algebra but also in calculus and beyond. A beneficial exercise improvement advice for students is to repeatedly practice factorization techniques to recognize patterns quickly.

Linear Equations

Linear equations are foundational to algebra and represent relationships where each variable is raised to at most the first power. A simple example is \( ax + b = 0 \), where \(a \) and \(b \) are constants. In partial fraction decomposition, once we have established our fractions with unknown constants (such as \( A \) and \( B \)), we often end up with a set of linear equations to solve and find the values of these constants.

As seen in the exercise, we set each extracted linear equation equal to a value determined by the original fraction's numerator. We then solved for \(A \) and \(B \) by plugging in specific values for \( x \) that make each equation true (like choosing values that cause one term to drop out). This systematic approach is a practical application of linear equations where solving for unknowns is straightforward, usually involving simple algebraic manipulations.

Understanding linear equations and their solutions are particularly important as they are not only relevant in algebra but also in more complex mathematics, and they form the basis of understanding how to manipulate and solve for variables in many real-world problems.