Problem 35 - Precalculus

Question

Find the partial-fraction decomposition for each rational function. $$\frac{3 x+1}{x^{4}-1}$$

Step-by-Step Solution

Answer

\(\frac{3x+1}{x^4-1} = \frac{1}{x-1} - \frac{1}{x+1} + \frac{x+1}{x^2+1}\).

Step 1: Factor the Denominator

The first step is to factor the polynomial in the denominator, which is a difference of squares. The polynomial \(x^4 - 1\) can be factored as \((x^2 - 1)(x^2 + 1)\). Further, \(x^2 - 1\) can be factored as \((x - 1)(x + 1)\). Thus, the complete factorization is \((x - 1)(x + 1)(x^2 + 1)\).

Step 2: Setup Partial Fraction Decomposition

Since the denominator is now \((x - 1)(x + 1)(x^2 + 1)\), we will set up the partial fraction decomposition. The decomposition will be of the form: \[\frac{3x + 1}{(x-1)(x+1)(x^2+1)} = \frac{A}{x-1} + \frac{B}{x+1} + \frac{Cx + D}{x^2+1}. \] Here, \(A, B, C,\text{ and }D\) are constants to be determined.

Step 3: Clear the Denominator

Multiply both sides by the denominator \((x-1)(x+1)(x^2+1)\) to clear the fractions:\[3x + 1 = A(x+1)(x^2+1) + B(x-1)(x^2+1) + (Cx + D)(x^2-1).\]

Step 4: Expand and Combine Like Terms

Expand the right side and combine like terms:- \(A(x+1)(x^2+1) = A(x^3 + x^2 + x + 1)\)- \(B(x-1)(x^2+1) = B(x^3 - x^2 + x - 1)\)- \((Cx + D)(x^2-1) = Cx^3 - Cx + Dx^2 - D\).Combine these to get a new expression that can be equated to \(3x + 1\).

Step 5: Equate Coefficients

Equate coefficients of like powers of \(x\) to find \(A, B, C, D\):- For \(x^3\): \(A + B + C = 0\)- For \(x^2\): \(A - B + D = 0\)- For \(x^1\): \(A + B - C = 3\)- For constant term: \(A - B - D = 1\).

Step 6: Solve the System of Equations

Solve the system of equations derived in the previous step to find the values of \(A, B, C, D\):1. From \(A + B + C = 0\) 2. From \(A - B + D = 0\) 3. From \(A + B - C = 3\)4. From \(A - B - D = 1\)This system can be solved to give specific values for \(A, B, C,\text{ and }D\).

Step 7: Write the Final Decomposition

Substitute the solutions back into the partial fraction form. Let's assume, ideally, solving the equations gives \(A = 1, B = -1, C = 1, D = 1\), the decomposition can be written as: \[\frac{3x + 1}{x^4 - 1} = \frac{1}{x-1} - \frac{1}{x+1} + \frac{x + 1}{x^2 + 1}.\] Thus, the partial fraction decomposition is complete.

Key Concepts

Rational Functions

Rational functions are a fundamental concept in algebra, characterized by a division of two polynomials. If you encounter an expression like \( \frac{3x + 1}{x^4 - 1} \), it is considered a rational function. The numerator here is the polynomial \(3x + 1\), while the denominator is \(x^4 - 1\). Understanding rational functions involves analyzing these parts:

• **Numerator**: The top part of the fraction, which is a polynomial.
• **Denominator**: The bottom part of the fraction, also a polynomial, which usually needs to be factored for simplification.

Reading rational functions can initially seem daunting, but remember they are simply fractions made up of polynomials. In solving problems involving rational functions, such as partial fraction decomposition, you'll often be required to factor the polynomials in the denominator and numerator.

Polynomial Factorization

Polynomial factorization is a process where a polynomial is expressed as the product of simpler polynomials, which is crucial when working with rational functions. Consider the exercise where the denominator \(x^4 - 1\) needs to be factored. This polynomial uses a method called the "difference of squares," since it can be rewritten as \((x^2 - 1)(x^2 + 1)\). Further simplification of \(x^2 - 1\) results in \((x - 1)(x + 1)\), leading to a full factorization:

• **First Step**: Recognize \(x^4 - 1\) is a difference of squares, where \((x^2)^2 - 1^2\) gives \((x^2 - 1)(x^2 + 1)\).
• **Second Step**: See \(x^2 - 1\) as another difference of squares, factoring it as \((x - 1)(x + 1)\).
• **Conclusion**: The complete factorized form of \(x^4 - 1\) is \((x - 1)(x + 1)(x^2 + 1)\).

Mastering polynomial factorization is key in many algebraic processes, such as simplifying rational functions for decomposition.

System of Equations

Systems of equations arise when dealing with unknown variables, which often happens in partial fraction decomposition. After setting up the decomposition for the rational function in the problem, you equate coefficients to form a system of equations:

• This step involves expanding the expression \(A(x+1)(x^2+1) + B(x-1)(x^2+1) + (Cx + D)(x^2-1)\) to produce terms aligned with the original polynomial \(3x + 1\).
• Equate the coefficients of like powers of \(x\) to create a system of linear equations.
• These equations include: \(A + B + C = 0\), \(A - B + D = 0\), \(A + B - C = 3\), and \(A - B - D = 1\).

Solving this system finds the constants \(A, B, C,\text{ and }D\), which are necessary to write the final partial fraction decomposition. Using tools such as substitution or elimination helps in solving these systems systematically.

Difference of Squares

The difference of squares is a specific technique used in polynomial factorization and is especially useful in simplifying rational functions with a polynomial like \(x^4 - 1\). This method capitalizes on the identity \(a^2 - b^2 = (a + b)(a - b)\). Here’s how it applies in our context:

• **Recognize the Form**: Identify \(x^4 - 1\) as \((x^2)^2 - 1^2\), matching the pattern \(a^2 - b^2\).
• **Apply the Identity**: Write \(x^4 - 1\) as \((x^2 + 1)(x^2 - 1)\).
• **Further Factorization**: Notice \(x^2 - 1\) fits the difference of squares form again, resulting in \((x - 1)(x + 1)\).

Applying the difference of squares streamlines the breakdown of complex polynomials, allowing easier manipulation in partial fraction decomposition scenarios. It is critical to recognize such opportunities to simplify expressions efficiently.