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Chapter 8: Problem 19

Write the partial fraction decomposition of each rational expression. $$\frac{2}{2 x^{2}-x}$$

Short Answer

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

Step by step solution

01

Simplify the rational function

First, factor the denominator of the rational function \(\frac{2}{2x^2-x}\) to get \(\frac{2}{x(2x-1)}\).
02

Write in terms of partial fractions

Then, write the simplified fraction using partial fraction decomposition as: \(\frac{2}{x(2x-1)} = \frac{A}{x} + \frac{B}{2x-1}\). The goal is to find A and B.
03

Clear the fractions

Multiply through by \(x(2x-1)\) (the denominator of the original fraction), which helps to clear the fractions. This can be written as: \(2 = A(2x-1) + Bx\).
04

Equating coefficients and solving

Now, it's necessary to equate the coefficients on each side of the equation to solve for A and B. Setting x = 0 gives \(A = -2\). Similarly, setting x = 1/2 yields \(B = 2\).

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Key Concepts

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

Understanding Rational Expressions
Rational expressions are fractions in which both the numerator and the denominator are polynomials. In the given exercise, \(\frac{2}{2x^2-x}\), the numerator is the constant 2, and the denominator is the quadratic polynomial \(2x^2-x\). To work with rational expressions, we often need to simplify them or break them down into simpler parts.

This can involve factoring the denominator, as seen in the exercise, where the denominator is factored into \(x(2x-1)\). Simplifying these expressions makes them easier to manage, particularly when integrating or looking for limits in calculus. A common technique for simplifying a rational expression is partial fraction decomposition, which is the focus of the exercise.
Equating Coefficients
Equating coefficients is a valuable technique used to find unknowns within an algebraic equation. For instance, when we have an equation such as \(2 = A(2x-1) + Bx\), we can find the coefficients A and B by making sure that the coefficients of corresponding powers of x on both sides of the equation match.

One approach is to substitute strategic values for x to isolate the coefficients. In our solution, setting \(x = 0\) simplifies the equation to \(2 = -2A\), hence \(A = -1\). Similarly, setting \(x = \frac{1}{2}\) yields \(2 = B\). Another approach would be to equate the coefficients of like terms on both sides, which would result in a system of linear equations where \(A\) and \(B\) can be determined.
Factoring Polynomials
Factoring polynomials is a critical skill that simplifies many types of algebraic tasks including solving equations, graphing functions, and as with our example, performing partial fraction decomposition. When factoring the denominator \(2x^2-x\), we look for common factors first. Here, we can pull out the common factor of \(x\), resulting in \(2x^2-x = x(2x-1)\).

The ability to recognize and factor common terms is essential in simplifying expressions and solving equations. Being proficient in different factoring techniques such as factoring by grouping, factoring perfect square trinomials, and factoring differences of squares, greatly enhances a student's algebraic toolbox.

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Most popular questions from this chapter

If \(A=\left[\begin{array}{ll}2 & 1 \\ 1 & 3\end{array}\right]\) and \(B=\left[\begin{array}{cc}2 & 2 a+b \\ b-a & 6\end{array}\right],\) for what values of \(a\) and \(b\) does \(A B=B A ?\)

A supply function for widgets is modeled by \(P(q)=a q+b,\) where \(q\) is the number of widgets supplied and \(P(q)\) is the total price of \(q\) widgets, in dollars. It is known that 200 widgets can be supplied for \(\$ 40\) and 100 widgets can be supplied for \(\$ 25 .\) Use a system of linear equations to find the constants \(a\) and \(b\) in the expression for the supply function.

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