Partial Fraction Decomposition: (2x^2-8x-2)/(x^3-x)
In this article, we will find the partial fraction decomposition of the rational expression (2x^2 - 8x - 2) / (x^3 - x). Partial fraction decomposition is a technique used to break down a complex rational expression into simpler fractions, making it easier to integrate, differentiate, or otherwise manipulate. This method is especially useful in calculus and engineering applications. Let's dive in!
1. Factor the Denominator
The first step in partial fraction decomposition is to factor the denominator of the rational expression. In our case, the denominator is x^3 - x. We can factor out an x:
x^3 - x = x(x^2 - 1)
Now, we recognize that x^2 - 1 is a difference of squares, which can be factored further:
x^2 - 1 = (x - 1)(x + 1)
So, the fully factored denominator is:
x^3 - x = x(x - 1)(x + 1)
Factoring the denominator is a critical initial step because it determines the form of the partial fractions. Each factor in the denominator will correspond to a term in the partial fraction decomposition. This foundational step sets the stage for the rest of the process, influencing how we set up and solve for the unknown constants. Understanding and executing this step accurately is essential for successful decomposition.
When the denominator is factored completely, we know we have accounted for all possible linear factors. In this specific problem, factoring the denominator into x(x-1)(x+1) allows us to see clearly that we need three partial fractions, each with one of these factors as its denominator. Without proper factorization, the subsequent steps would be based on an incorrect foundation, leading to a wrong final answer. Thus, meticulous attention to detail during factorization is paramount. This is particularly true when dealing with more complex polynomials in the denominator, where the factorization might not be immediately obvious and may require techniques like synthetic division or more advanced algebraic manipulation. The factored form allows us to express the original rational function as a sum of simpler fractions, which is the core idea behind partial fraction decomposition.
2. Set Up the Partial Fraction Decomposition
Now that we have the factored denominator, we can set up the partial fraction decomposition. Since we have three distinct linear factors, we will have three partial fractions, each with one of these factors as its denominator. The numerators will be constants, which we will call A, B, and C.
(2x^2 - 8x - 2) / (x(x - 1)(x + 1)) = A/x + B/(x - 1) + C/(x + 1)
This equation represents the decomposition of the original rational expression into simpler fractions. Our next goal is to find the values of A, B, and C.
Setting up the partial fraction decomposition correctly is paramount for solving the problem. The structure of the decomposition directly depends on the factors of the denominator. Each unique linear factor corresponds to a partial fraction with a constant numerator. Repeated linear factors would require additional terms with increasing powers of the factor in the denominator. For instance, if we had a factor like (x-2)^2, we would need two terms: A/(x-2) + B/(x-2)^2. Similarly, irreducible quadratic factors in the denominator require linear expressions in the numerator (e.g., (Ax + B)/(x^2 + 1)).
The setup must accurately reflect the nature of the denominator's factors to ensure the subsequent steps are valid. Failing to account for repeated or irreducible quadratic factors will lead to an incorrect decomposition. In our case, with distinct linear factors, the setup is straightforward. We assign a constant numerator to each factor, which simplifies the algebraic manipulations needed to solve for the constants. The equation we create at this stage is the foundation upon which we build our solution. A clear and accurate setup translates to a smoother and more reliable path to finding the values of A, B, and C. Therefore, it's crucial to double-check the factors and ensure each is properly represented in the decomposition.
3. Clear the Fractions
To solve for A, B, and C, we need to clear the fractions. We do this by multiplying both sides of the equation by the original denominator, which is x(x - 1)(x + 1):
(2x^2 - 8x - 2) = A(x - 1)(x + 1) + Bx(x + 1) + Cx(x - 1)
This step eliminates the denominators and gives us a polynomial equation that we can solve for the unknown constants.
Clearing the fractions is a pivotal step in simplifying the equation and enabling us to solve for the unknown constants. This process involves multiplying both sides of the equation by the least common denominator (LCD), which in this case is x(x - 1)(x + 1). By doing so, we eliminate the fractional terms, converting the equation into a polynomial equation that is much easier to manipulate.
When multiplying each term by the LCD, make sure to distribute correctly and cancel out common factors in each fraction. This step requires careful attention to detail to avoid errors, as any mistake here will propagate through the rest of the solution. After clearing the fractions, the resulting polynomial equation relates the original numerator to the constants A, B, and C, multiplied by various polynomial terms. This equation forms the basis for solving the constants, either by substituting specific values of x or by equating coefficients of like terms. Either approach aims to create a system of linear equations that can be solved for the values of A, B, and C, which are essential for completing the partial fraction decomposition. Therefore, accuracy and precision in clearing the fractions are crucial for obtaining the correct decomposition.
4. Solve for A, B, and C
Now we can solve for A, B, and C. There are a couple of methods we can use. One method is to substitute specific values of x that will eliminate some of the terms. Another method is to expand the equation and equate coefficients.
Method 1: Substituting Values of x
Let's start by substituting x = 0 into the equation:
2(0)^2 - 8(0) - 2 = A(0 - 1)(0 + 1) + B(0)(0 + 1) + C(0)(0 - 1)
-2 = A(-1)(1) + 0 + 0
-2 = -A
A = 2
Next, let's substitute x = 1 into the equation:
2(1)^2 - 8(1) - 2 = A(1 - 1)(1 + 1) + B(1)(1 + 1) + C(1)(1 - 1)
2 - 8 - 2 = 0 + B(1)(2) + 0
-8 = 2B
B = -4
Finally, let's substitute x = -1 into the equation:
2(-1)^2 - 8(-1) - 2 = A(-1 - 1)(-1 + 1) + B(-1)(-1 + 1) + C(-1)(-1 - 1)
2 + 8 - 2 = 0 + 0 + C(-1)(-2)
8 = 2C
C = 4
So, we have found A = 2, B = -4, and C = 4.
Solving for the constants is the core of the partial fraction decomposition process. The method of substituting values of x is particularly effective when dealing with distinct linear factors, as it allows us to strategically eliminate terms and solve for each constant individually. By choosing values of x that make one or more factors zero, we can isolate the remaining constant and solve for its value directly. This approach simplifies the algebraic manipulations and reduces the risk of errors.
For instance, in our example, substituting x = 0 eliminated the terms with B and C, allowing us to solve for A directly. Similarly, substituting x = 1 and x = -1 allowed us to solve for B and C, respectively. This method works because each chosen value of x simplifies the equation to a point where only one unknown constant remains. This not only makes the calculations easier but also provides a clear and direct path to finding the values of the constants. However, it is important to select appropriate values of x to avoid division by zero or undefined expressions, which could lead to incorrect results. By carefully choosing the values of x, we can efficiently and accurately determine the constants needed for the partial fraction decomposition.
Method 2: Equating Coefficients
Alternatively, we can expand the equation and equate coefficients. Starting from the equation:
2x^2 - 8x - 2 = A(x - 1)(x + 1) + Bx(x + 1) + Cx(x - 1)
Expand the terms:
2x^2 - 8x - 2 = A(x^2 - 1) + B(x^2 + x) + C(x^2 - x)
2x^2 - 8x - 2 = Ax^2 - A + Bx^2 + Bx + Cx^2 - Cx
Now, group the terms by powers of x:
2x^2 - 8x - 2 = (A + B + C)x^2 + (B - C)x - A
Equate the coefficients:
For x^2: A + B + C = 2
For x: B - C = -8
For the constant term: -A = -2, so A = 2
Now we can substitute A = 2 into the first equation:
2 + B + C = 2
B + C = 0
We have two equations with two unknowns:
B + C = 0
B - C = -8
Add the two equations:
2B = -8
B = -4
Substitute B = -4 into B + C = 0:
-4 + C = 0
C = 4
So, we have found A = 2, B = -4, and C = 4, which matches our previous results.
Equating coefficients is an alternative method for solving the constants in partial fraction decomposition, particularly useful when the denominator contains higher-degree polynomials or repeated factors. This approach involves expanding all terms in the equation after clearing fractions and then grouping terms with the same powers of x. By equating the coefficients of corresponding powers of x on both sides of the equation, we create a system of linear equations that can be solved for the unknown constants.
For instance, in our example, expanding and grouping terms allowed us to create three equations: A + B + C = 2, B - C = -8, and -A = -2. Solving this system of equations gives us the values of A, B, and C. This method is especially valuable when substituting values of x is not straightforward or does not provide enough equations to solve for all constants. While equating coefficients can involve more algebraic manipulation compared to substitution, it provides a systematic way to determine the constants. However, it is essential to ensure accuracy in expanding and grouping terms to avoid errors in the resulting system of equations. By carefully applying this method, we can successfully determine the constants needed for the partial fraction decomposition.
5. Write the Partial Fraction Decomposition
Now that we have found A, B, and C, we can write the partial fraction decomposition:
(2x^2 - 8x - 2) / (x^3 - x) = 2/x - 4/(x - 1) + 4/(x + 1)
This is the partial fraction decomposition of the given rational expression.
Writing the partial fraction decomposition is the final step in the process, where we substitute the values of the constants (A, B, and C) back into the original setup. This step consolidates all the previous work, providing the decomposed form of the rational expression. Accuracy in this step is crucial, as any error in substituting the constants will lead to an incorrect final result. The final decomposition expresses the original complex rational function as a sum of simpler fractions, each with a constant numerator and a linear denominator.
In our example, after finding A = 2, B = -4, and C = 4, we substitute these values back into the equation (2x^2 - 8x - 2) / (x^3 - x) = A/x + B/(x - 1) + C/(x + 1) to obtain the final decomposition: (2x^2 - 8x - 2) / (x^3 - x) = 2/x - 4/(x - 1) + 4/(x + 1). This decomposition is now ready for various applications, such as integration, differentiation, or other algebraic manipulations. It is always a good practice to double-check the final result by recombining the partial fractions to see if they simplify back to the original expression. By carefully executing this final step, we ensure that the partial fraction decomposition is complete and accurate.
Conclusion
We have successfully found the partial fraction decomposition of the rational expression (2x^2 - 8x - 2) / (x^3 - x). The decomposition is:
(2x^2 - 8x - 2) / (x^3 - x) = 2/x - 4/(x - 1) + 4/(x + 1)
This technique is valuable in calculus and other areas of mathematics. Remember to factor the denominator, set up the partial fractions, clear the fractions, solve for the constants, and write the final decomposition.
For further reading on partial fraction decomposition, you can visit Paul's Online Math Notes.
