Problem 56
Question
Find the partial fraction expansion. \(\frac{x^{3}-4 x^{2}+5 x+4}{(x-2)^{3}}\)
Step-by-Step Solution
Verified Answer
The partial fraction expansion is \(\frac{1}{x-2} + \frac{9}{(x-2)^2} + \frac{22}{(x-2)^3}.\)
1Step 1: Understanding the Problem
The task is to find the partial fraction expansion of the given rational function \(\frac{x^{3}-4 x^{2}+5 x+4}{(x-2)^{3}}\). This means we need to express it as a sum of simpler fractions.
2Step 2: Set Up the Partial Fraction Decomposition
The denominator \((x-2)^3\) suggests that the partial fraction form should involve terms of the form \(\frac{A}{x-2} + \frac{B}{(x-2)^2} + \frac{C}{(x-2)^3}\), where \(A\), \(B\), and \(C\) are constants we need to determine.
3Step 3: Write the Equation
The equation to solve is: \[\frac{x^{3}-4 x^{2}+5 x+4}{(x-2)^{3}} = \frac{A}{x-2} + \frac{B}{(x-2)^2} + \frac{C}{(x-2)^3}.\] Multiply through by \((x-2)^3\) to get: \[x^{3}-4 x^{2}+5 x+4 = A(x-2)^2 + B(x-2) + C.\]
4Step 4: Expand and Group Terms
Expand \(A(x-2)^2\) and \(B(x-2)\). \[A(x^2 - 4x + 4) = Ax^2 - 4Ax + 4A\] \[B(x-2) = Bx - 2B\] Combine these with \(C\) to get: \[Ax^2 - 4Ax + 4A + Bx - 2B + C.\]
5Step 5: Equate and Solve for Constants
Match the coefficients from both sides of the equation: 1. For \(x^3\), there is no \(x^3\) term on the right, but on the left it's \(x^3\). Thus, \(A = 1.\) 2. For \(x^2\), on the left, it is \(-4x^2\). On the right, it is \(0x^2\), so this gives us no new equation. 3. For \(x\), the coefficients are \(5\) (left) and \((-4A + B)x\) (right). Substitute \(A = 1\), then \(-4(1) + B = 5\), so \(B = 9.\) 4. The constant terms provide: \(4 = 4A - 2B + C\), substituting in \(A = 1\) and \(B = 9\), \(4 = 4(1) - 2(9) + C\), solving gives \(C = 22.\)
6Step 6: Write the Final Partial Fraction Expansion
We now know the values of \(A\), \(B\), and \(C\). Substitute them back into the partial fraction layout: \[\frac{x^{3}-4 x^{2}+5 x+4}{(x-2)^{3}} = \frac{1}{x-2} + \frac{9}{(x-2)^2} + \frac{22}{(x-2)^3}.\]
Key Concepts
Rational FunctionsPolynomial DivisionAlgebraic FractionsEquation Solving
Rational Functions
A rational function is essentially a fraction where both the numerator and the denominator are polynomials. For example, \(\frac{x^{3}-4 x^{2}+5 x+4}{(x-2)^{3}}\) is a rational function. The key components to look at are the degree of the polynomial in the numerator and the degree of the polynomial in the denominator. By examining these degrees, we can determine the behavior of the rational function at different points.
Understanding rational functions is fundamental to learning partial fraction decomposition, as it involves breaking down the complicated fraction into simpler, more manageable parts.
Understanding rational functions is fundamental to learning partial fraction decomposition, as it involves breaking down the complicated fraction into simpler, more manageable parts.
Polynomial Division
Polynomial division is a process similar to long division with numbers. It is often used to simplify rational functions or to prepare them for further operations, like partial fraction decomposition.
In this context, sometimes it may be necessary to divide the polynomials first, especially if the degree of the numerator is higher than that of the denominator. This process reduces the degree of the expression, making it easier to handle.
In this context, sometimes it may be necessary to divide the polynomials first, especially if the degree of the numerator is higher than that of the denominator. This process reduces the degree of the expression, making it easier to handle.
- Just like in arithmetic division, you'll divide the leading term of the numerator by the leading term of the denominator.
- Then, multiply and subtract to find the remainder, continuing the process until you've broken down the expression.
Algebraic Fractions
Algebraic fractions are fractions that contain variables. When working with these, it's important to understand how to manipulate them to solve equations or simplify expressions.
Partial fraction decomposition is one technique that helps us express an algebraic fraction as a sum of simpler fractions. This is particularly useful when dealing with expressions where the denominator is factorable.
Partial fraction decomposition is one technique that helps us express an algebraic fraction as a sum of simpler fractions. This is particularly useful when dealing with expressions where the denominator is factorable.
- First, factor the denominator if possible.
- Then express your fraction as a sum of fractions with constant numerators.
Equation Solving
Solving equations involving rational functions often requires multiple steps, including simplifying the expression, partial fraction decomposition, and solving for unknowns.
In the given exercise, we express a complicated fraction into simpler parts, each with its own fraction. We solve for the constants by equating coefficients from our expanded expression.
In the given exercise, we express a complicated fraction into simpler parts, each with its own fraction. We solve for the constants by equating coefficients from our expanded expression.
- Identify the terms that need to be equated from both sides of the equation.
- Solve for each constant individually.
- Plug these constants back into the original expression to verify the solution matches the original rational function.
Other exercises in this chapter
Problem 56
Your roommate, John, offered to buy household supplies for you and your other roommate. You live near the border of three states, each of which has a different
View solution Problem 56
For the following exercises, construct a system of nonlinear equations to describe the given behavior, then solve for the requested solutions. The squares of tw
View solution Problem 56
Solve for the desired quantity. A stuffed animal business has a total cost of production \(C=12 x+30\) and a revenue function \(R=20 x .\) Find the break-even p
View solution Problem 56
For the following exercises, solve for the desired quantity. A stuffed animal business has a total cost of production \(C=12 x+30\) and a revenue function \(R=2
View solution