Problem 13
Question
Writing the Form of the Decomposition. Write the form of the partial fraction decomposition of the rational expression. Do not solve for the constants. $$\frac{4 x^{2}+3}{(x-5)^{3}}$$
Step-by-Step Solution
Verified Answer
The partial fraction decomposition of the rational expression \(\frac{4 x^{2}+3}{(x-5)^{3}}\) is \(\frac{A}{(x-5)^{3}} + \frac{B}{(x-5)^{2}} + \frac{C}{(x-5)}\). Where \(A\), \(B\), and \(C\) are constants.
1Step 1: Identify the Form of the Denominator
The denominator in the expression is \((x-5)^{3}\). As the power of \((x-5)\) is 3, there will be three fractions in the decomposition. As the power is more than 1, we need to write down decreasing powers of \((x-5)\) starting from 3.
2Step 2: Formulate Partial Fractions
There should be as many fractions as the highest power of the denominator. Hence, start with the highest power of 3 and decreasing until 1, each respective fraction should look like these: \(\frac{A}{(x-5)^{3}}\), where \(A\) is the constant for the first fraction. \(\frac{B}{(x-5)^{2}}\), where \(B\) is the constant for the second fraction. \(\frac{C}{(x-5)}\), where \(C\) is the constant for the third fraction.
3Step 3: Write the Completed Partial Fraction Decomposition
Now, combine all the fractions from Step 2 to show the complete partial fraction decomposition. Therefore, it becomes: \(\frac{4 x^{2}+3}{(x-5)^{3}} =\frac{A}{(x-5)^{3}} + \frac{B}{(x-5)^{2}} + \frac{C}{(x-5)}\)
Key Concepts
Rational ExpressionsPowers of PolynomialsConstants in Partial Fractions
Rational Expressions
Rational expressions are expressions that involve a ratio or fraction of two polynomials. These are in the form \( \frac{P(x)}{Q(x)} \), where \( P(x) \) and \( Q(x) \) are polynomials and \( Q(x) eq 0 \).
The numerator and the denominator are polynomials. In rational expressions, it's crucial the denominator is not zero since division by zero is undefined.
These expressions can be simplified or decomposed into simpler partial fractions, which help in integration and other operations in calculus.
When dealing with rational expressions, especially in the context of partial fraction decomposition, the focus is often on the polynomial in the denominator.
The numerator and the denominator are polynomials. In rational expressions, it's crucial the denominator is not zero since division by zero is undefined.
These expressions can be simplified or decomposed into simpler partial fractions, which help in integration and other operations in calculus.
When dealing with rational expressions, especially in the context of partial fraction decomposition, the focus is often on the polynomial in the denominator.
- It is necessary to assess the degree of the polynomial.
- This degree determines the method used for decomposition.
Powers of Polynomials
Polynomials, such as \( (x-5)^3 \), involve variables raised to powers. The power or exponent signifies how many times the base, in this case \( x-5 \), is multiplied by itself.
For example, \( (x-5)^3 \) is equivalent to \((x-5) \cdot (x-5) \cdot (x-5)\).
Higher powers indicate more complex behavior in graphs and equations, leading to a more layered approach in decomposition. The process involves:
For example, \( (x-5)^3 \) is equivalent to \((x-5) \cdot (x-5) \cdot (x-5)\).
Higher powers indicate more complex behavior in graphs and equations, leading to a more layered approach in decomposition. The process involves:
- Breaking down the powers sequentially in decreasing order.
- Starting with the highest power in the denominator.
Constants in Partial Fractions
When performing partial fraction decomposition, the constants \( A \), \( B \), and \( C \) are essential parts of the formula. They represent unknown values that are effectively placeholders in the expression.
These constants ensure each term of the original rational expression is represented correctly.
These constants ensure each term of the original rational expression is represented correctly.
- Each constant corresponds to a fraction whose denominator is a power of the original polynomial.
- They are crucial for reconstruction of the original expression when combined.
Other exercises in this chapter
Problem 13
Solving a System by Elimination In Exercises \(13-30\) , solve the system by the method of elimination and check any solutions algebraically. $$ \left\\{\begin{
View solution Problem 13
Solving a System by Substitution In Exercises \(7-14,\) solve the system by the method of substitution. Check your solution(s) graphically. $$\left\\{\begin{arr
View solution Problem 13
Using Back-Substitution In Exercises \(11 - 16 ,\) use back-substitution to solve the system of linear equations. $$\left\\{ \begin{aligned} 2 x + y - 3 z & = 1
View solution Problem 14
Solving a Linear Programming Problem, sketch the region determined by the constraints. Then find the minimum and maximum values of the objective function (if po
View solution