Problem 45
Question
In Exercises \(41-46,\) find the constants \(A, B,\) and \(C\). $$\frac{5 x^{2}+1}{(x+1)\left(x^{2}-x+1\right)}=\frac{A}{x+1}+\frac{B x+C}{x^{2}-x+1}$$
Step-by-Step Solution
Verified Answer
#Answer#
The decomposition of the given rational function is:
$$
\frac{5x^2+1}{(x+1)(x^2-x+1)} = \frac{1/2}{x+1}+\frac{(9/2)x+1/2}{x^2-x+1}
$$
1Step 1: Clear the denominators
Multiply both sides of the equation by \((x+1)(x^2-x+1)\):
$$
5x^2+1=A(x^2-x+1)+(Bx+C)(x+1)
$$
2Step 2: Group the terms
Expand the equation and group the terms:
$$
5x^2+1=Ax^2-Ax+A+Bx^2+xB+Cx+C
$$
Combine the like terms and rewrite the equation:
$$
5x^2+1=(A+B)x^2+(C-A)x+(A+C)
$$
3Step 3: Solve for A, B, and C
Compare the coefficients of the powers of x:
$$
\begin{cases}
A+B=5\\
C-A=0\\
A+C=1
\end{cases}
$$
4Step 4: Solve the system of equations
Solving the system of equations, we get:
$$
\begin{cases}
A+B=5\\
A=C\\
A+C=1
\end{cases}
\Rightarrow
\begin{cases}
A=1/2\\
C=1/2\\
B=9/2
\end{cases}
$$
Therefore, the decomposition of the given rational function is:
$$
\frac{5x^2+1}{(x+1)(x^2-x+1)} = \frac{1/2}{x+1}+\frac{(9/2)x+1/2}{x^2-x+1}
$$
Key Concepts
Rational FunctionsSystems of EquationsPolynomial Long Division
Rational Functions
Rational functions are mathematical expressions that represent the ratio of two polynomials. They have the form \( \frac{P(x)}{Q(x)} \) where \( P(x) \) and \( Q(x) \) are polynomials and \( Q(x) \) is not equal to zero.
In our exercise example, we have the rational function \( \frac{5x^2+1}{(x+1)(x^2-x+1)} \). It's important to understand that rational functions can often be simplified or decomposed into simpler parts known as partial fractions. This decomposition assists with integration and other calculus operations.
To decompose a rational function, we typically look for a set of simpler fractions with unknown constants that sum up to the original function. The exercise you’re tackling involves finding these constants (\( A, B, C \) in this case) so that when the decomposition is reassembled, it equals the original rational function. Discovering these constants requires a firm grasp of polynomial functions and their behavior.
In our exercise example, we have the rational function \( \frac{5x^2+1}{(x+1)(x^2-x+1)} \). It's important to understand that rational functions can often be simplified or decomposed into simpler parts known as partial fractions. This decomposition assists with integration and other calculus operations.
To decompose a rational function, we typically look for a set of simpler fractions with unknown constants that sum up to the original function. The exercise you’re tackling involves finding these constants (\( A, B, C \) in this case) so that when the decomposition is reassembled, it equals the original rational function. Discovering these constants requires a firm grasp of polynomial functions and their behavior.
Systems of Equations
A system of equations is a collection of two or more equations with a shared set of unknowns. In the situation with our rational function, after clearing the denominators and expanding, we obtain a system of linear equations that we must solve in order to find the constants \( A, B, \) and \( C \).
The system derived from the exercise is a set of equations with common variables, each equating to various coefficients of the original expression. It is represented by:
\[ \begin{cases} A+B=5\ C-A=0\ A+C=1 \end{cases}\]
To solve this system, we can use a variety of methods, including substitution, elimination, or matrix operations. In this case, the step-by-step solution utilized substitution, plugging the value of \( A \) from one equation into the others to find the values of \( B \) and \( C \). It's evident that an understanding of how to manipulate and solve systems of equations is crucial for successfully performing partial fraction decomposition.
The system derived from the exercise is a set of equations with common variables, each equating to various coefficients of the original expression. It is represented by:
\[ \begin{cases} A+B=5\ C-A=0\ A+C=1 \end{cases}\]
To solve this system, we can use a variety of methods, including substitution, elimination, or matrix operations. In this case, the step-by-step solution utilized substitution, plugging the value of \( A \) from one equation into the others to find the values of \( B \) and \( C \). It's evident that an understanding of how to manipulate and solve systems of equations is crucial for successfully performing partial fraction decomposition.
Polynomial Long Division
Polynomial long division is a method used to divide a polynomial by another polynomial of lower degree, much like long division with numbers. While not directly used in this step-by-step solution, it is a useful tool to keep in mind, especially when working with rational functions.
Here's a quick overview: if in a given rational expression, the degree of the numerator is higher than the denominator, before doing partial fraction decomposition, we should use polynomial long division to divide the polynomials. This process simplifies the function into a form that can be more easily decomposed into partial fractions.
To perform polynomial long division, you divide the terms of the highest degree first and continue the process until the remainder's degree is lower than that of the divisor. The quotient becomes part of the simplified expression, and any remainder is written as a rational function (if it exists). Although in our exercise, the degrees in the numerator and the denominator suggest that direct decomposition is suitable, it's good to be aware of this technique for more complex cases.
Here's a quick overview: if in a given rational expression, the degree of the numerator is higher than the denominator, before doing partial fraction decomposition, we should use polynomial long division to divide the polynomials. This process simplifies the function into a form that can be more easily decomposed into partial fractions.
To perform polynomial long division, you divide the terms of the highest degree first and continue the process until the remainder's degree is lower than that of the divisor. The quotient becomes part of the simplified expression, and any remainder is written as a rational function (if it exists). Although in our exercise, the degrees in the numerator and the denominator suggest that direct decomposition is suitable, it's good to be aware of this technique for more complex cases.
Other exercises in this chapter
Problem 44
In Exercises \(41-46,\) find the constants \(A, B,\) and \(C\). $$\frac{x^{2}-x-21}{(2 x-1)\left(x^{2}+4\right)}=\frac{A}{2 x-1}+\frac{B x+C}{x^{2}+4}$$
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In Exercises \(41-46,\) find the constants \(A, B,\) and \(C\). $$\frac{x-2}{(x+4)\left(x^{2}+2 x+2\right)}=\frac{A}{x+4}+\frac{B x+C}{x^{2}+2 x+2}$$
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Lillian borrows \(\$ 10,000 .\) She borrows some from her friend at \(8 \%\) annual interest, twice as much as that from her bank at \(9 \%,\) and the remainder
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