Chapter 26: Problem 613
Decompose \(\left(\mathrm{x}^{2}+4 \mathrm{x}\right) /\left[(\mathrm{x}-2)^{2}\left(\mathrm{x}^{2}+4\right)\right]\) into partial fractions.
Short Answer
Expert verified
The partial fraction decomposition of \(\frac{x^2 + 4x}{(x-2)^2(x^2+4)}\) is given by \(\frac{3}{2(x-2)^2}+\frac{x+24}{12(x^2+4)}\).
Step by step solution
01
Identify the partial fraction decomposition structure
Given the rational function
\[
\frac{x^2 + 4x}{(x-2)^2(x^2+4)}
\]
We will decompose it into a sum of simpler rational functions. Since the denominator has a square of a linear term, we need an additional term in the decomposition compared to the case with distinct linear factors. Thus, the decomposition has the following form:
\[
\frac{x^2 + 4x}{(x-2)^2(x^2+4)} = \frac{A}{x-2} + \frac{B}{(x-2)^2} + \frac{Cx+D}{x^2+4}
\]
02
Clear the denominators
To find the values of constants A, B, C, and D, multiply both sides of the equation by the denominator to clear the fractions:
\[
x^2 + 4x = A(x-2)^2(x^2+4) + B(x^2+4) + (Cx+D)(x-2)^2
\]
03
Plug in x = 2
To simplify the expression, we can plug in \(x = 2\) on both sides, which sets the first and third terms on the right-hand side to 0, giving:
\[
(2)^2 + 4(2) = B((2)^2+4)
\]
Solving for B, we get:
\[
B = \frac{12}{8} = \frac{3}{2}
\]
04
Expanding and equating coefficients
Now, we want to expand the right-hand side of the equation and collect like terms:
\[
x^2 + 4x = A[(x^2+4)(x-2)^2]+ \frac{3}{2}(x^2+4) + (Cx+D)(x-2)^2
\]
Expanding the terms, we get:
\[
x^2 + 4x =x^4(2A) - 8x^3 A + (24Ax^2 + 2Cx^2) + (-32Ax + 4Dx) + (16A - 8C+ 4D)
\]
Now, we equate the coefficients of like terms on both sides of the equation. We have:
\[
\begin{cases}
2A = 0 \\
-8A = 0 \\
24A + 2C = 1 \\
-32A + 4D = 4 \\
16A - 8C + 4D = 0
\end{cases}
\]
05
Solve the system of equations
Solving the system of equations, we find:
\[
A = 0, \quad B = \frac{3}{2}, \quad C = \frac{1}{12}, \quad D = 2
\]
06
Write the partial fraction decomposition
Now that we have the values of A, B, C, and D, we can write the final partial fraction decomposition of the given rational function:
\[
\frac{x^2 + 4x}{(x-2)^2(x^2+4)} = \frac{0}{x-2} + \frac{\frac{3}{2}}{(x-2)^2} + \frac{\frac{1}{12}x+2}{x^2+4} = \frac{3}{2(x-2)^2} + \frac{x+24}{12(x^2+4)}
\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Rational Functions
Rational functions are expressions that involve a ratio of two polynomials. In simpler terms, they are fractions where both the numerator and denominator are polynomials. For the given exercise, the rational function is:
- Numerator: \(x^2 + 4x\)
- Denominator: \((x-2)^2(x^2+4)\)
Algebraic Expressions
Algebraic expressions, like \(x^2 + 4x\), consist of numbers, variables, and operations. The numerator in our problem, \(x^2 + 4x\), is a quadratic expression, which means it has a degree of two because the highest power of \(x\) is 2. The denominator \((x-2)^2(x^2+4)\) is a product of linear and quadratic terms.
Algebraic expressions can usually be factored and manipulated to reveal their simpler components. These manipulations allow us to break down more complex expressions into partial fractions. It’s crucial to understand how each component of the expression can affect the overall behavior of the rational function, particularly in terms of where it is undefined (due to the denominator equal to zero) or simplified. Understanding algebraic expressions and factoring them correctly is vital in the partial decomposition process.
Algebraic expressions can usually be factored and manipulated to reveal their simpler components. These manipulations allow us to break down more complex expressions into partial fractions. It’s crucial to understand how each component of the expression can affect the overall behavior of the rational function, particularly in terms of where it is undefined (due to the denominator equal to zero) or simplified. Understanding algebraic expressions and factoring them correctly is vital in the partial decomposition process.
Equating Coefficients
Equating coefficients is a key step in arranging and solving algebraic equations, especially when dealing with partial fraction decomposition. Once we set up our decomposition equation, the next move is to expand, as shown in the exercise:
After substituting values like \(x = 2\) to find constants, we expanded further:
\[x^2 + 4x = A[(x^2+4)(x-2)^2]+ \frac{3}{2}(x^2+4) + (Cx+D)(x-2)^2\]
This results in an equation where we align like terms and equate coefficients to solve for unknowns like \(A\), \(B\), \(C\), and \(D\). By comparing terms of the same degree on both sides of the equation, we form a system of equations that can be solved to determine these constants. This process is pivotal because it allows us to break down a complex rational function into simpler expressions that are much easier to work with.
After substituting values like \(x = 2\) to find constants, we expanded further:
\[x^2 + 4x = A[(x^2+4)(x-2)^2]+ \frac{3}{2}(x^2+4) + (Cx+D)(x-2)^2\]
This results in an equation where we align like terms and equate coefficients to solve for unknowns like \(A\), \(B\), \(C\), and \(D\). By comparing terms of the same degree on both sides of the equation, we form a system of equations that can be solved to determine these constants. This process is pivotal because it allows us to break down a complex rational function into simpler expressions that are much easier to work with.
System of Equations
A system of equations arises when we equate coefficients in partial fraction decomposition. In this context, solving a system means finding the values of unknown variables (\(A\), \(B\), \(C\), \(D\)) that satisfy all equations simultaneously.
In the exercise:
In the exercise:
- \(2A = 0\)
- \(-8A = 0\)
- \(24A + 2C = 1\)
- \(-32A + 4D = 4\)
- \(16A - 8C + 4D = 0\)