Problem 43
Question
Find the partial fraction decomposition of the rational function. $$\frac{x^{5}-2 x^{4}+x^{3}+x+5}{x^{3}-2 x^{2}+x-2}$$
Step-by-Step Solution
Verified Answer
Use polynomial long division, then express as a sum of the quotient and partial fractions of the remainder.
1Step 1: Check If Proper Rational Function
A rational function is proper if the degree of the numerator is less than the degree of the denominator. Here, the numerator has degree 5 and the denominator has degree 3. Since the degree of the numerator is greater, it is an improper fraction.
2Step 2: Perform Polynomial Long Division
Divide \(x^5 - 2x^4 + x^3 + x + 5\) by \(x^3 - 2x^2 + x - 2\). The first term is \(x^2\). Multiply the entire divisor by \(x^2\) and subtract from the original polynomial. Repeat the process with the resulting polynomial. After division, you should find the quotient and remainder.
3Step 3: Express the Original Function
After division, let the quotient be \(Q(x)\) and the remainder be \(R(x)\). Then, express the original rational function as: \[ Q(x) + \frac{R(x)}{x^3 - 2x^2 + x - 2} \]
4Step 4: Factor the Denominator
The denominator \(x^3 - 2x^2 + x - 2\) may need to be factored further if possible to identify simple factors or irreducible quadratic factors. Use synthetic division or the Rational Root Theorem to find factors, making it simpler to decompose.
5Step 5: Set Up Partial Fractions
Assuming the denominator factors into linear or irreducible quadratic terms, set up the partial fractions accordingly. For instance, if it factors into \((x-1)(x-1)(x+1)\), set up partial fractions like: \[ \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x+1} \]
6Step 6: Solve for Coefficients
To solve for coefficients, multiply both sides by the factored denominator to clear the fractions. Compare coefficients on both sides or plug in suitable values for \(x\) to create a system of equations to solve for \(A\), \(B\), and \(C\).
7Step 7: Write the Decomposition
Substitute all found coefficients back into the partial fractions. Combine this with any quotient found in Step 3 for the complete decomposed expression.
Key Concepts
Polynomial Long DivisionRational FunctionImproper FractionSynthetic Division
Polynomial Long Division
Polynomial Long Division is a fundamental method used to divide polynomials, much like how long division is used with numbers.
This process comes into play when dealing with dividing a polynomial by another polynomial.
It helps us simplify complex expressions and is especially useful when dealing with improper fractions in rational functions.
Here's how it works in a simple way:
This process comes into play when dealing with dividing a polynomial by another polynomial.
It helps us simplify complex expressions and is especially useful when dealing with improper fractions in rational functions.
Here's how it works in a simple way:
- Set up the division: Write the dividend (the polynomial to be divided) and the divisor (the polynomial doing the dividing) as you would with regular numbers.
- Figure out how many times the leading term of the divisor fits into the leading term of the dividend.
Use the result as the first term of your quotient. Repeat the process. - Multiply the entire divisor by this new term, write the result under the dividend, and subtract to find the new dividend.
- Continue this until each term of the dividend has been accounted for.
Rational Function
A Rational Function is a fraction where both the numerator and the denominator are polynomials.
It is expressed in the form: \[ f(x) = \frac{P(x)}{Q(x)} \]
where \(P(x)\) and \(Q(x)\) are polynomials and \(Q(x) eq 0\).
Some important points about rational functions include:
It is expressed in the form: \[ f(x) = \frac{P(x)}{Q(x)} \]
where \(P(x)\) and \(Q(x)\) are polynomials and \(Q(x) eq 0\).
Some important points about rational functions include:
- The domain of a rational function is all real numbers except where the denominator equals zero.
- A rational function is called 'proper' when the degree of the polynomial in the numerator is less than the degree in the denominator.
- It is 'improper' when the numerator's degree is greater than or equal to the denominator's degree, requiring division to simplify.
Improper Fraction
An Improper Fraction in the context of rational functions is when the degree of the numerator is at least as large as that of the denominator.
For example, in the exercise, the numerator has a degree of 5 while the denominator has a degree of 3, making it an improper fraction.
Here’s why it matters:
For example, in the exercise, the numerator has a degree of 5 while the denominator has a degree of 3, making it an improper fraction.
Here’s why it matters:
- You need to convert an improper fraction into a proper fraction.
This helps in further analysis, such as partial fraction decomposition. - The conversion involves using polynomial long division to find a quotient and a remainder.
- After division, the result expresses the improper fraction as a sum of the quotient and a proper fraction.
Synthetic Division
Synthetic Division is a simplified form of polynomial division that can be easier and faster than polynomial long division.
It is mainly used when dividing by a linear factor of the form \((x - c)\).
This method is efficient for finding the roots of polynomials and is often used when attempting to factor a polynomial.
Here's how it works in simple terms:
It is mainly used when dividing by a linear factor of the form \((x - c)\).
This method is efficient for finding the roots of polynomials and is often used when attempting to factor a polynomial.
Here's how it works in simple terms:
- Identify the root \(c\) from the linear factor \((x - c)\).
- Set up a table that includes the coefficients of the polynomial you wish to divide.
- Drop the leading coefficient down, multiply by \(c\), and add to the next coefficient. Repeat this across the row.
- The last value in your computed row represents the remainder, while the other values form the coefficients of the quotient polynomial.
Other exercises in this chapter
Problem 43
Let $$\begin{array}{l} A=\left[\begin{array}{cccc} 1 & 0 & 6 & -1 \\ 2 & \frac{1}{2} & 4 & 0 \end{array}\right] \quad C=\left[\begin{array}{r} 1 \\ 0 \\ -1 \\ -
View solution Problem 43
Solve the system of linear equations. $$\left\\{\begin{array}{rr} x+2 y-3 z= & -5 \\ -2 x-4 y-6 z= & 10 \\ 3 x+7 y-2 z= & -13 \end{array}\right.$$
View solution Problem 43
Solve the system, or show that it has no solution. If the system has infinitely many solutions, express them in the ordered pair form given in Example 6. $$\lef
View solution Problem 44
Find the inverse of the matrix. For what value(s) of \(x\) if any, does the matrix have no inverse? $$\left[\begin{array}{cc} e^{x} & -e^{2 x} \\ e^{2 x} & e^{3
View solution