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Chapter 12: Problem 21

Find the partial fraction decomposition of each rational expression. $$ \frac{x}{(x-1)(x-2)} $$

Short Answer

Expert verified
\frac{x}{(x-1)(x-2)} = \frac{-1}{x-1} + \frac{2}{x-2}

Step by step solution

01

Set up the partial fraction decomposition

Write the given rational expression as a sum of partial fractions. Assume:\[\frac{x}{(x-1)(x-2)} = \frac{A}{x-1} + \frac{B}{x-2}\]where A and B are constants to be determined.
02

Clear the denominators

Multiply both sides of the equation by the common denominator (x-1)(x-2):\[x = A(x-2) + B(x-1)\]
03

Expand and collect like terms

Expand the right-hand side:\[x = Ax - 2A + Bx - B\]Combine like terms:\[x = (A + B)x - (2A + B)\]
04

Set up the system of equations

To find A and B, equate the coefficients of corresponding terms from both sides of the equation. This gives us two equations:1. For the coefficient of x: \[1 = A + B\]2. For the constant term: \[0 = -2A - B\]
05

Solve the system of equations

Solve the system of equations to find the values of A and B. From 1: \[A + B = 1\]. From 2: \[-2A - B = 0\]. Solve equation 2 for B: \[B = -2A\]. Substitute B in equation 1: \[A - 2A = 1\] which simplifies to \[-A = 1\]. Thus, \[A = -1\]. Substitute A back into \[B = -2A\] to get \[B = 2\].
06

Write the partial fraction decomposition

Using the found values of A and B, write the partial fraction decomposition:\[\frac{x}{(x-1)(x-2)} = \frac{-1}{x-1} + \frac{2}{x-2}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Rational Expressions
Rational expressions are fractions where the numerator and the denominator are polynomials. For example, \(\frac{x}{(x-1)(x-2)}\) in our exercise. Simplifying and decomposing these expressions helps us understand their behavior and simplifies many algebraic operations.

When working with rational expressions, always look to factor denominators completely and identify simpler partial fractions that sum back to the original expression. This makes analysis and computation easier.
System of Equations
To determine the constants in partial fraction decomposition, we set up a system of equations. This involves equating coefficients of like terms on both sides of our fraction conversion equation.

In our example, by comparing coefficients of x and the constant terms, we derived these equations: \[1 = A + B\] and \[0 = -2A - B\]. Solving these equations yields values for A and B, which fit the original rational expression.

This approach ensures that our decomposition accurately represents the original fraction.
Constants Determination
Determining the constants like A and B in partial fraction decomposition is crucial. Begin by setting up your fraction and clearing the denominator via multiplication. After expanding, match coefficients of like terms.

In this case: \[x = A(x-2) + B(x-1)\] expands to \[x = (A + B)x - 2A - B\], matching the original polynomial by solving: \[1 = A + B\] and \[0 = -2A - B\]. This logical setup helps accurately find values for A and B. Use substitution methods to isolate and solve for these constants.
Common Denominator
To start partial fraction decomposition, we first find a common denominator linking all fractions involved. Here, \((x-1)(x-2)\) serves as the unifying factor. Clearing denominators involves multiplying both sides by this common term.

For example, multiplying both sides of \(\frac{x}{(x-1)(x-2)} = \frac{A}{x-1} + \frac{B}{x-2}\) leads to: \[x = A(x-2) + B(x-1)\]. This prepares the equation for expansion and matching coefficients, crucial steps to finding the unknowns A and B. This method roots our solution in algebraic fundamentals, making it robust and reliable.

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Most popular questions from this chapter

Write a brief paragraph outlining your strategy for solving a system of two linear equations containing two variables.

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