Question

Resolve \(\frac{2 x+1}{x^{2}-2 x-8}\) into partial fractions. (1) \(\frac{3}{2(x-4)}-\frac{1}{2(x+2)}\) (2) \(\frac{3}{2(x-4)}+\frac{1}{2(x+2)}\) (3) \(\frac{1}{2(x-4)}-\frac{3}{2(x+2)}\) (4) None of these

Short answer

Question: Decompose the given expression into partial fractions: $$\frac{2x+1}{x^2-2x-8}$$ a) $$\frac{3}{2(x-4)} - \frac{1}{2(x+2)}$$ b) $$\frac{1}{x-4} - \frac{3}{x+2}$$ c) $$\frac{2}{x-4} + \frac{1}{x+2}$$ d) $$\frac{1}{2(x-4)} + \frac{3}{2(x+2)}$$ Answer: a) $$\frac{3}{2(x-4)} - \frac{1}{2(x+2)}$$

Step-by-step solution

Factor the denominator

Factor the denominator expression, \(x^2-2x-8\), into its factors. This can be done by solving a quadratic equation: \((x - 4)(x + 2)\). So, our factored expression is \((x - 4)(x + 2)\).

Decompose into partial fractions

We can now write the fraction as the sum of two partial fractions. Let's assume the constants A and B for the numerators of the partial fractions. $$\frac{2x+1}{(x-4)(x+2)} = \frac{A}{x-4} + \frac{B}{x+2}$$

Clear the denominators and compare coefficients

To find the constants A and B, we need to eliminate the denominators and compare the coefficients. Start by multiplying both sides with the common denominator: $$(2x+1) = A(x+2) + B(x-4)$$ Now, compare the coefficients of the terms on both sides of the equation: - For x's coefficient: \(2 = A + B\) - For constant term: \(1 = 2A - 4B\)

Solve the system of equations for A and B

Now, solve the system of equations for A and B: $$\begin{cases} A + B = 2\\ 2A -4B = 1 \end{cases}$$ Multiplying the first equation by 2, we get \(2A + 2B = 4\). Now, subtract the second equation from this new equation: $$6B = 3 \Rightarrow B = \frac{1}{2}$$ Now, substitute the value of B into the first equation: $$A + \frac{1}{2} = 2 \Rightarrow A = \frac{3}{2}$$

Write the partial fractions with the found coefficients

Now that we have the values for A and B, we can write the decomposition of the given fraction into partial fractions: $$\frac{2x+1}{(x-4)(x+2)} = \frac{3}{2(x-4)} - \frac{1}{2(x+2)}$$ Thus, the correct answer is option (1).

Key concepts

Mathematical Decomposition

Mathematical decomposition refers to the process of breaking down complex mathematical expressions into simpler, more manageable parts. This technique is especially useful when dealing with integrals, differential equations, and algebraic equations. In the context of algebraic fractions, it involves expressing a complicated fraction as a sum or difference of simpler fractions, known as 'partial fractions'.

This process is akin to taking a complex puzzle and separating it into smaller pieces that are easier to handle and understand. Just as with a puzzle, once the pieces are understood individually, they can be put back together to solve the original, complex problem. Mathematical decomposition not only makes calculations more straightforward but also gives insight into the structure of the original expression, which can be invaluable in further mathematical analysis.

Algebraic Fractions

What Are Algebraic Fractions?

Algebraic fractions are fractions that contain algebraic expressions in their numerators, denominators, or both. They follow the same principles as arithmetic fractions but involve variables. Solving algebraic fractions often requires finding a common denominator or breaking them into partial fractions, as seen in the provided exercise.

To simplify algebraic fractions, factoring the expressions in the denominators is a critical step because it reveals the simpler fractions that could combine to form the original expression. Mastery of working with algebraic fractions is essential for advanced algebra, calculus, and beyond, where they frequently arise in more complex situations.

Quadratic Equations

Breaking Down Quadratic Equations

Quadratic equations are polynomial equations of degree two. The general form is \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(a\) is not equal to zero. These equations are foundational in algebra and appear in various applications across mathematics and the sciences.

The process of solving quadratic equations often involves factoring, completing the square, using the quadratic formula, or graphing. In the context of partial fractions decomposition, factoring a quadratic is essential—it allows us to express the denominator as a product of two simpler terms which can then be used to decompose the fraction into partial fractions.

System of Equations

Solving a System of Equations

A system of equations consists of multiple equations that share common variables. The goal is to find the values of these variables that satisfy all equations in the system simultaneously. Methods for solving systems include graphing, substitution, elimination, and matrices.

In the context of partial fractions, we use the system of equations to determine the unknown constants that are part of the decomposed fractions. By comparing coefficients and solving the system of equations, we can find these constants and thus complete the decomposition. This step is critical: without solving the system, we cannot identify the unique values that allow us to reformulate the original complex fraction into its simpler components.