Problem 3
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. $$\frac{x^{2}-3 x+5}{(x-2)^{2}(x+4)}$$
Step-by-Step Solution
Verified Answer
\( \frac{A}{x-2} + \frac{B}{(x-2)^2} + \frac{C}{x+4} \)
1Step 1: Identify the Denominator Factors
The denominator of the given fraction is \( (x-2)^{2}(x+4) \). The factors are \( (x-2) \) repeated twice and \( (x+4) \) once.
2Step 2: Set Up Partial Fraction Form
Since \( (x-2) \) is a repeated linear factor, include terms for each power up to the repeat count. Include terms for each distinct linear factor:\[ \frac{x^2 - 3x + 5}{(x-2)^{2}(x+4)} = \frac{A}{x-2} + \frac{B}{(x-2)^2} + \frac{C}{x+4} \] Here, \( A \), \( B \), and \( C \) are the coefficients to be determined.
Key Concepts
Rational FunctionsPolynomial FunctionsCoefficients in Algebra
Rational Functions
Understanding rational functions is crucial for tackling problems like partial fraction decomposition. A rational function is essentially a ratio of two polynomial functions. In this case, you have a fraction where the numerator and denominator are polynomial expressions. Rational functions are extensively used because they capture a wide range of behaviors with relatively simple expressions.
The given function, \( \frac{x^2 - 3x + 5}{(x-2)^2(x+4)} \), is a classic example of a rational function.
The given function, \( \frac{x^2 - 3x + 5}{(x-2)^2(x+4)} \), is a classic example of a rational function.
- The numerator is a polynomial of degree 2: \( x^2 - 3x + 5 \).
- The denominator, however, is more complex because it involves products of polynomial factors.
Polynomial Functions
Polynomial functions are the building blocks of rational functions. They are expressions consisting of variables raised to whole number powers, multiplied by coefficients. Understanding polynomial functions is key to unraveling their behavior and methods of manipulation, such as factorization and decomposition.
This exercise involves a polynomial expression in the numerator \( x^2 - 3x + 5 \), and the denominator \( (x-2)^2(x+4) \) involves two polynomial factors.
This exercise involves a polynomial expression in the numerator \( x^2 - 3x + 5 \), and the denominator \( (x-2)^2(x+4) \) involves two polynomial factors.
- The numerator is a linear combination of terms, giving it a degree of 2.
- The denominator, meanwhile, is composed of a squared term \((x-2)^2\) and a linear term \((x+4)\), introducing a required structured approach for decomposition.
Coefficients in Algebra
Coefficients are the numerical factors that multiply the variables in an algebraic expression. In partial fraction decomposition, finding these coefficients is often the last and critical step. Here, you focus on determining coefficients that will break down the rational function into simpler fractions.
In the solution setup \( \frac{A}{x-2} + \frac{B}{(x-2)^2} + \frac{C}{x+4} \), \( A \), \( B \), and \( C \) are the coefficients that must be determined to ensure that when summed, these fractions reconstruct the original function.
In the solution setup \( \frac{A}{x-2} + \frac{B}{(x-2)^2} + \frac{C}{x+4} \), \( A \), \( B \), and \( C \) are the coefficients that must be determined to ensure that when summed, these fractions reconstruct the original function.
- These coefficients play a pivotal role in defining the "weights" of each partial fraction.
- Determining them requires setting up equations based on equating the original function with the decomposed form.
Other exercises in this chapter
Problem 3
Find the determinant of the matrix, if it exists. $$\left[\begin{array}{rr} 4 & 5 \\ 0 & -1 \end{array}\right]$$
View solution Problem 3
State whether the equation or system of equations is linear. $$\left\\{\begin{aligned} x y-3 y+z &=5 \\ x-y^{2}+5 z &=0 \\ 2 x+y z &=3 \end{aligned}\right.$$
View solution Problem 3
Graph the inequality. $$y>x$$
View solution Problem 3
Calculate the products \(A B\) and \(B A\) to verify that \(B\) is the inverse of \(A\). $$A=\left[\begin{array}{rrr} 1 & 3 & -1 \\ 1 & 4 & 0 \\ -1 & -3 & 2 \en
View solution