Problem 18 - Algebra and Trigonometry

Question

Find the partial fraction decomposition of the rational function. \(\frac{2 x+1}{x^{2}+x-2}\)

Step-by-Step Solution

Answer

The partial fraction decomposition is \( \frac{1}{x+2} + \frac{1}{x-1} \).

Step 1: Factor the Denominator

The first step in finding the partial fraction decomposition of a rational function is to factor the denominator. The denominator is \(x^2 + x - 2\). We need to find two numbers that multiply to \(-2\) and add to \(1\). These numbers are \(2\) and \(-1\). Thus, we can factor the quadratic expression as \((x+2)(x-1)\).

Step 2: Set Up the Partial Fraction Decomposition

With the denominator factored as \((x+2)(x-1)\), we can set up the partial fraction decomposition. Since both of these factors are linear, the decomposition will have the form:\[ \frac{2x + 1}{x^2 + x - 2} = \frac{A}{x+2} + \frac{B}{x-1} \]where \(A\) and \(B\) are constants that we need to determine.

Step 3: Clear the Denominator and Combine Fractions

Multiply both sides of the equation by the common denominator \((x+2)(x-1)\) to clear the fractions. This results in:\[ 2x + 1 = A(x-1) + B(x+2) \]Simplify the right-hand side:\[ = Ax - A + Bx + 2B \]This combines to:\[ (A + B)x + (2B - A) \]

Step 4: Equate Coefficients

Now equate the coefficients from both sides of the equation:- The coefficient of \(x\) gives: \( A + B = 2 \)- The constant term gives: \( 2B - A = 1 \)We now have a system of two linear equations to solve for \(A\) and \(B\).

Step 5: Solve the System of Equations

Solve the system:1. \( A + B = 2 \)2. \( 2B - A = 1 \)From equation (1): \( A = 2 - B \). Substitute into equation (2):\[ 2B - (2 - B) = 1 \]Simplify:\[ 2B - 2 + B = 1 \]\[ 3B = 3 \]\[ B = 1 \]Substitute back to find \(A\):\[ A = 2 - 1 = 1 \]

Step 6: Write the Partial Fraction Decomposition

With \(A = 1\) and \(B = 1\), substitute these values back into the partial fraction decomposition:\[ \frac{2x + 1}{x^2 + x - 2} = \frac{1}{x+2} + \frac{1}{x-1} \]This is the final partial fraction decomposition of the given rational function.

Key Concepts

Rational Functions

A rational function is essentially a fraction where both the numerator and denominator are polynomials. In the expression \( \frac{2x + 1}{x^2 + x - 2} \), the numerator \( 2x + 1 \) is a linear expression, and the denominator \( x^2 + x - 2 \) is a quadratic expression. Understanding these types of functions is key in algebra because they often appear in various problems and solutions.
A major characteristic of rational functions is that they can potentially have undefined values, specifically where the denominator equals zero. Hence, solving them usually involves simplifying the expression, finding values that nullify the denominator, and then analyzing the function's behavior.
Partial fraction decomposition is a technique used to break down complex rational functions into simpler fractions, which can then be used for further analysis or integration.

Factoring Quadratics

Factoring quadratics involves rewriting a quadratic expression, such as \( x^2 + x - 2 \), into a product of linear factors. This means you are looking for two binomials that, when multiplied together, give you the original quadratic.
To factor the quadratic \( x^2 + x - 2 \), we need numbers that multiply to \(-2\) (the constant term) and add up to \(1\) (the coefficient of the \(x\) term). The numbers \(2\) and \(-1\) work perfectly for this, allowing us to write the expression as \((x+2)(x-1)\).

• This transformation is crucial for simplifying expressions and solving equations.
• It also helps in setting up the initial form of a partial fraction decomposition.

Factoring plays a big role in reducing complex algebra problems into more manageable ones.

Linear Equations

Linear equations are equations of the first degree, meaning they involve variables raised only to the first power. When breaking down complex problems, like rational functions, identifying linear relationships through these equations becomes vital.
In partial fraction decomposition, once the expression is set up as \( \frac{A}{x+2} + \frac{B}{x-1} \), solving for \(A\) and \(B\) involves creating and solving linear equations.
Equating the coefficients of \(x\) and the constant terms results in the following system:

• \(A + B = 2\)
• \(2B - A = 1\)

These are linear equations that need solving to find the values of \(A\) and \(B\), which are crucial for completing the partial fraction decomposition.

System of Equations

A system of equations consists of two or more equations that share the same set of variables, and the goal is to find values for these variables that satisfy all the equations simultaneously.
In the context of partial fraction decomposition, we solve the system:

• \(A + B = 2\)
• \(2B - A = 1\)

Here, we are tasked with solving for \(A\) and \(B\). By substituting \(A = 2 - B\) from the first equation into the second, you simplify the problem into one equation with one variable:
\[3B = 3\]
This yields \(B = 1\), and substituting back gives \(A = 1\). Systems of equations are a vital algebraic tool for breaking down and solving parts of more complex problems like the one we have here.