Question

Find \(a, b,\) and \(c\) such that $$ \frac{x^{2}-x+3}{\left(x^{2}+2\right)(2 x-1)}=\frac{a x+b}{x^{2}+2}+\frac{c}{2 x-1} $$

Short answer

\(a = 0\), \(b = -1\), \(c = 1\)

Step-by-step solution

Identify the Structure

We are given a rational function on the left side and two separate fractions on the right side. Our task is to find constants \(a\), \(b\), and \(c\) such that both sides are equivalent.

Multiply Both Sides by Common Denominator

Multiply both sides of the equation by the denominator \((x^2 + 2)(2x - 1)\), to eliminate the denominators:\[(x^2 - x + 3) = (ax + b)(2x - 1) + c(x^2 + 2)\]

Expand and Simplify

Distribute and expand the right side to remove the parentheses:\[(x^2 - x + 3) = (2ax + b)x - (ax + b) + cx^2 + 2c\]

Combine Like Terms

Combine all terms on the right-hand side by organizing them into powers of \(x\): \[ x^2(2a + c) + x(b - a) + (-b + 2c) \]

Compare Coefficients

Equate the coefficients from both sides of the equation for \(x^2\), \(x\), and the constant term:- For \(x^2\): \(2a + c = 1\)- For \(x\): \(b - a = -1\)- For the constant term: \(-b + 2c = 3\)

Solve the System of Equations

Solve the resulting system of linear equations: 1. \(2a + c = 1\)2. \(b - a = -1\)3. \(-b + 2c = 3\). From equation 2, we express \(b = a - 1\). Substitute \(b\) in the third equation and solve:\[-(a-1) + 2c = 3 \Rightarrow -a + 1 + 2c = 3 \Rightarrow -a + 2c = 2\]We now have a simplified system:\[ \begin{align*} 2a + c &= 1 \ -a + 2c &= 2 \end{align*}\]Solve this system by substitution or elimination methods to find \(a = 0\), \(c = 1\), and \(b\) based on \(b = a - 1\), thus \(b = -1\).

Verify the Solution

Substitute \(a = 0\), \(b = -1\), \(c = 1\) back into the original equation to confirm the left and right sides are equal after simplification.

Key concepts

Rational Functions

Rational functions are expressions that represent the ratio of two polynomials. In simpler terms, they look like fractions with polynomials on top, termed as the numerator, and on the bottom, termed as the denominator. The degree of the polynomial in the numerator and the denominator plays a key role in understanding the behavior of the rational function. In our exercise, the rational function given is \( \frac{x^{2}-x+3}{(x^{2}+2)(2x-1)} \). Here, \(x^{2}-x+3\) is the polynomial numerator, while \((x^{2}+2)(2x-1)\) forms the polynomial denominator.
When dealing with rational functions, one common goal is to break them down into simpler parts using a method called partial fraction decomposition. This involves expressing the complex rational function as a sum of simpler fractions, each with a well-defined denominator. This decomposition is particularly useful in calculus for simplifying integrations and helps in solving complex mathematical equations. So the task at hand is to decompose \( \frac{x^{2}-x+3}{(x^{2}+2)(2x-1)} \) into simpler parts that can be handled more easily, allowing for a more detailed understanding and solving of related problems.

Linear Equations

Linear equations form the bedrock of algebra and are characterized by expressions where the highest power of the variable is one. In the exercise, the decomposition of the rational function ultimately leads us to a system of linear equations that look like \(2a + c = 1\), \(b - a = -1\), and \(-b + 2c = 3\).

Solving Systems of Linear Equations

Manipulating these equations is key to solving for unknowns like \(a\), \(b\), and \(c\). These systems can often be solved using substitution or elimination methods:

• Substitution involves expressing one variable in terms of others and replacing it in the remaining equations.
• Elimination focuses on adding or subtracting equations to remove one of the variables, thus simplifying the system.

For instance, in this problem, we express \(b = a - 1\) by rearranging the second equation. Then, substitute this into the third equation to express it only in terms of \(a\) and \(c\). Finally, using elimination or substitution, you find \(a = 0\), \(c = 1\), and then \(b = -1\). These techniques are essential and widely applicable in various mathematics disciplines.

Coefficient Comparison

In the context of partial fraction decomposition, comparing coefficients is a vital step. This technique is used to equate the coefficients of powers of variables on both sides of the equation, which helps in identifying the unknown parameters. After clearing the denominators by multiplying with the common denominator, we expand both sides of the equation.

Logic Behind Coefficient Comparison

Consider the expanded form \((ax + b)(2x - 1) + c(x^2 + 2)\) which matches with \(x^2 - x + 3\) after simplification. By comparing the coefficients of \(x^2\), \(x\), and the constant term, we derive relationships involving \(a\), \(b\), and \(c\):

• For \(x^2\), the coefficient is \(2a + c = 1\).
• For \(x\), the coefficient is \(b - a = -1\).
• For the constant term, \(-b + 2c = 3\).

Identifying these equations allows us to form a system of linear equations that can then be solved. This step not only helps in confirming the decomposition's correctness but also in pinpointing the exact values of the constants involved.