Problem 9
Question
Write the form of the partial fraction decomposition of the function (as in Example 4). Do not determine the numerical values of the coefficients. \(\frac{x^{3}+x+1}{x(2 x-5)^{3}\left(x^{2}+2 x+5\right)^{2}}\)
Step-by-Step Solution
Verified Answer
\(\frac{A}{x} + \frac{B}{2x-5} + \frac{C}{(2x-5)^2} + \frac{D}{(2x-5)^3} + \frac{Ex + F}{x^2 + 2x + 5} + \frac{Gx + H}{(x^2 + 2x + 5)^2}\).
1Step 1: Analyze the Denominator
Identify the different types of factors in the denominator: we have the linear factor \(x\), the repeated linear factor \((2x-5)^3\), and the repeated irreducible quadratic factor \((x^2 + 2x + 5)^2\).
2Step 2: Set Up Partial Fractions for Linear Factors
For the linear factor \(x\), assign a term \(\frac{A}{x}\) where \(A\) is a constant to be determined.
3Step 3: Set Up Partial Fractions for Repeated Linear Factor
For the repeated linear factor \((2x-5)^3\), assign terms for each power up to 3: \(\frac{B}{2x-5} + \frac{C}{(2x-5)^2} + \frac{D}{(2x-5)^3}\).
4Step 4: Set Up Partial Fractions for Repeated Irreducible Quadratic Factor
For \((x^2 + 2x + 5)^2\), assign terms in the form of \(\frac{Ex + F}{x^2 + 2x + 5}\) and \(\frac{Gx + H}{(x^2 + 2x + 5)^2}\).
5Step 5: Combine All Parts
Combine all terms together to form the full expression for the partial fraction decomposition: \(\frac{A}{x} + \frac{B}{2x-5} + \frac{C}{(2x-5)^2} + \frac{D}{(2x-5)^3} + \frac{Ex + F}{x^2 + 2x + 5} + \frac{Gx + H}{(x^2 + 2x + 5)^2}\).
Key Concepts
Linear FactorsRepeated Linear FactorsIrreducible Quadratic FactorsRational Functions
Linear Factors
In the world of algebra and calculus, linear factors are a core concept of partial fraction decomposition. A linear factor is a polynomial of the first degree, meaning it can be expressed in the form of \(ax + b\), where \(a\) and \(b\) are constants, and \(a\) is not zero. In the given expression, the factor \(x\) is an example of a simple linear factor. When decomposing a rational function, each linear factor receives a fractional term consisting of a constant numerator and the linear factor itself as the denominator. For instance, for \(x\), we assign a term \(\frac{A}{x}\).
- This approach helps in simplifying complex rational expressions.
- It makes integration and other calculus operations more manageable.
Repeated Linear Factors
Repeated linear factors emerge in partial fraction decomposition when a linear factor is raised to a power greater than one. In our expression, \( (2x-5)^3 \) exemplifies this concept, where the linear factor \((2x-5)\) is repeated three times.To decompose such factors, we write separate terms for each power starting from one up to the factor's exponent. For \((2x-5)^3\), we assign:
- \(\frac{B}{2x-5}\)
- \(\frac{C}{(2x-5)^2}\)
- \(\frac{D}{(2x-5)^3}\)
Irreducible Quadratic Factors
Irreducible quadratic factors are polynomial expressions of degree two that do not have real roots. In other words, they can't be factored further into real linear factors. The term \((x^2 + 2x + 5)\) in our denominator is an example of an irreducible quadratic factor.For such factors in partial fraction decomposition, we use a numerator of the form \(Ax + B\) rather than a single constant. This accounts for any polynomial expression up to the degree of the quadratic factor. For example:
- \(\frac{Ex + F}{x^2 + 2x + 5}\)
- \(\frac{Gx + H}{(x^2 + 2x + 5)^2}\)
Rational Functions
Rational functions are fractions in which both the numerator and the denominator are polynomials. The study and breakdown of rational functions are central to understanding partial fraction decomposition, as seen in our expression \(\frac{x^{3}+x+1}{x(2 x-5)^{3}(x^{2}+2 x+5)^{2}}\).The goal of partial fraction decomposition is to simplify a complex rational function into simpler, easily manageable parts. This is achieved by breaking down the rational function based on its denominator factors, whether they are linear, repeated, or irreducible quadratic factors.
- This simplification aids in integrating rational functions, where direct integration might be difficult.
- It is also useful in solving differential equations.
Other exercises in this chapter
Problem 8
Use the substitution method to find all solutions of the system of equations. \(\left\\{\begin{aligned} x^{2}-y &=1 \\ 2 x^{2}+3 y &=17 \end{aligned}\right.\)
View solution Problem 9
1–14 Graph the inequality. $$4 x+5 y
View solution Problem 9
\(9-14\) Evaluate the minor and cofactor using the matrix \(A\) $$ A=\left[\begin{array}{rrr}{1} & {0} & {\frac{1}{2}} \\ {-3} & {5} & {2} \\\ {0} & {0} & {4}\e
View solution Problem 9
Find the inverse of the matrix if it exists. \(\left[\begin{array}{rr}{2} & {5} \\ {-5} & {-13}\end{array}\right]\)
View solution