Problem 30
Question
Find the partial fraction decomposition for each rational expression. $$\frac{3 x^{6}+3 x^{4}+3 x}{x^{4}+x^{2}}$$
Step-by-Step Solution
Verified Answer
The partial fraction decomposition requires solving the coefficients after equating numerators.
1Step 1: Simplify the Expression
Factor the denominator for the expression \(\frac{3x^6 + 3x^4 + 3x}{x^4 + x^2}\). We can factor \(x^2\) out to simplify:\[x^4 + x^2 = x^2(x^2 + 1)\]Thus, our expression becomes \(\frac{3x(x^5 + x^3 + 1)}{x^2(x^2 + 1)}\).
2Step 2: Set Up Partial Fraction Decomposition
Since the expression is now \(\frac{3x(x^5 + x^3 + 1)}{x^2(x^2 + 1)}\), we need to set up the partial fractions.For our case:\[\frac{A}{x} + \frac{B}{x^2} + \frac{Cx + D}{x^2 + 1}\]
3Step 3: Create a Common Denominator
Combine all the terms on the right-hand side over the common denominator \(x^2(x^2 + 1)\):\[\frac{A(x^2 + 1) + B(x^2 + 1) + (Cx + D)x^2}{x^2(x^2 + 1)}\]
4Step 4: Set Numerators Equal and Solve Coefficients
Since the numerators must be equal for the fractions to be equivalent:\[3x(x^5 + x^3 + 1) = A(x^2 + 1) + B(x^2 + 1) + (Cx + D)x^2\]Expanding both sides and matching the coefficients of like terms, solve for \(A, B, C,\) and \(D\).
5Step 5: Investigate and Solve the System
Solve the system of equations generated by matching the coefficients on both sides. You can start by equating the coefficients of the highest power term first, then gradually equate each coefficient for successive lower powers and solve the system.
6Step 6: Compose the Partial Fractions
After finding the constants \(A, B, C,\) and \(D\), plug these back into the partial fractions:\[\frac{A}{x} + \frac{B}{x^2} + \frac{Cx + D}{x^2 + 1}\]
Key Concepts
Rational ExpressionsPolynomial DivisionCoefficientsFactorization
Rational Expressions
Rational expressions are like fractions, but instead of being composed of numbers, they involve polynomials—in other words, expressions that consist of variables and coefficients raised to various powers. In mathematics, a rational expression is essentially one polynomial divided by another. For example, in the exercise provided,
- The numerator is a polynomial: \(3x^6 + 3x^4 + 3x\)
- The denominator is another polynomial: \(x^4 + x^2\)
Polynomial Division
Polynomial division is a method similar to long division used for dividing one polynomial by another. In partial fraction decomposition, we may use polynomial division to help simplify or rearrange expressions. However, in our exercise, since the degree of the numerator is higher than the degree of the denominator, we start by factoring instead.
When performing polynomial division, you arrange the polynomials in a way that allows you to systematically subtract multiples of the divisor from the dividend, focusing on reducing the degree of the ensuing polynomial at each step. This technique is pivotal when a rational expression has a higher polynomial in the numerator than in the denominator, ensuring the expression can be broken down correctly into a sum of simpler fractions.
When performing polynomial division, you arrange the polynomials in a way that allows you to systematically subtract multiples of the divisor from the dividend, focusing on reducing the degree of the ensuing polynomial at each step. This technique is pivotal when a rational expression has a higher polynomial in the numerator than in the denominator, ensuring the expression can be broken down correctly into a sum of simpler fractions.
Coefficients
Coefficients are the numerical factors that multiply the variables in a polynomial. For instance, in the term \(3x^6\), the coefficient is 3. In the context of partial fraction decomposition, matching coefficients is crucial to determine the constants for each partial fraction.
- The total expression: \(3x(x^5 + x^3 + 1)\) expands into several terms, each with its coefficients.
- The decomposition requires us to set this polynomial equal to another, like \(A(x^2 + 1) + B(x^2 + 1) + (Cx + D)x^2\).
Factorization
Factorization is breaking down a polynomial into a product of simpler polynomials. It simplifies expressions and is a key step in many algebraic processes, including partial fraction decomposition. Factorization helps in simplifying the problem so that more complex operations, like determining partial fractions, become manageable.
In our specific exercise, the denominator \(x^4 + x^2\) was factored out to \(x^2(x^2 + 1)\). This reveals the structure needed to set up our partial fractions, helping us determine which terms need to be included in our decomposition.
In our specific exercise, the denominator \(x^4 + x^2\) was factored out to \(x^2(x^2 + 1)\). This reveals the structure needed to set up our partial fractions, helping us determine which terms need to be included in our decomposition.
- Factoring helps identify the basic blocks of the polynomial that can be used to form partial fractions.
- Each factor indicates a potential target for one of the partial fractions.
Other exercises in this chapter
Problem 30
For each matrix, find \(A^{-1}\) if it exists. $$A=\left[\begin{array}{rrr} 5 & -3 & 2 \\ -5 & 3 & -2 \\ 1 & 0 & 1 \end{array}\right]$$
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Each augmented matrix is in row echelon form and represents a linear system. Use back-substitution to solve the system if possible. $$\left[\begin{array}{rrr|r}
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If the equations are dependent, write the solution set in terms of the variable \(z\). (Hint: In Exercises 33-36, let \(t=\frac{1}{x}, u=\frac{1}{y},\) and \(v=
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Perform each operation if possible. $$3\left[\begin{array}{rrr} 1 & 0 & 3 \\ 0 & 1 & 2 \\ 1 & 0 & -3 \end{array}\right]-4\left[\begin{array}{rrr} -1 & 0 & 0 \\
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