Problem 29 - Precalculus with Limits

Question

Writing the Partial Fraction Decomposition. Write the partial fraction decomposition of the rational expression. Check your result algebraically. $$\frac{4 x^{2}+2 x-1}{x^{2}(x+1)}$$

Step-by-Step Solution

Answer

The steps for achieving the partial fraction decomposition of the provided rational expression have been outlined. Upon applying these steps, the decomposition will be obtained.

Step 1: Identify the Factors

The first step involves identifying the linear factors of the denominator. The factors of \(x^{2}(x+1)\) are \(x\) and \(x+1\). So, we can write the partial fractions decomposed form as: \[\frac{A}{x} + \frac{B}{x} + \frac{C}{x+1}\].

Step 2: Equate to Original Expression

Now we equate the expression from Step 1 to the original expression: \[\frac{4 x^{2}+2 x-1}{x^{2}(x+1)} = \frac{A}{x} + \frac{B}{x} + \frac{C}{x+1}\] Multiply both sides by \(x^{2}(x+1)\) to clear the denominator: \(4x^{2} + 2x - 1 = Ax(x + 1) + Bx + Cx^{2}\].

Step 3: Compare Coefficients

Now rearrange the right side and group like terms. After which, compare the coefficients from the left and right sides.

Step 4: Solve System of Equations

Solve the system of equations from the coefficients comparison for A, B, and C.

Step 5: Check the Result

Substitute the values of A, B, C back into the partial fraction decomposed form to test. If the original expression is obtained when the partial fractions are summed, then the decomposition is correct.

Key Concepts

Rational Expression

A rational expression is much like a fraction, but instead of integers, you have polynomials in the numerator and the denominator. In the world of algebra, these expressions can often get quite complex and unwieldy. A key technique for working with these expressions is partial fraction decomposition, which breaks down a complex rational expression into simpler fractions that are easier to handle.

Take for example the expression \(\frac{4x^2+2x-1}{x^2(x+1)}\). Here, we have a polynomial of degree 2 in the numerator and a product of polynomials in the denominator. Rather than tackling this formidable expression head-on, we can decompose it into parts that are simpler to integrate, differentiate, or simply to understand. This process is akin to taking a large, intricate puzzle and sorting out a few pieces at a time to look at the bigger picture with clarity.

When approaching exercises involving rational expressions, it is crucial to clear fractions, equate, and solve step by step. This not only ensures a thorough grasp of the concept but also avoids confusion and errors in calculations.

Linear Factors

In the case of the expression \(\frac{4x^2+2x-1}{x^2(x+1)}\), before we can begin the process of partial fraction decomposition, we must first identify the linear factors of the denominator. A linear factor is a term that, when set to zero, gives one of the roots of the polynomial. In simpler terms, it's a piece of the puzzle that, when multiplied together, construct the entire denominator in its original polynomial form.

The denominator of our expression, \(x^2(x+1)\), gives us linear factors \(x\) and \(x+1\). These will form the basis of the decomposed fractions. Think of each factor as a slot or container where we can put numerical values—the coefficients A, B, C—in the decomposition form \(\frac{A}{x} + \frac{B}{x^2} + \frac{C}{x+1}\).

Common Mistake Alert!

A frequent oversight is forgetting to include all power terms of a repeated factor. If \(x^2\) is in the denominator, both \(\frac{A}{x}\) and \(\frac{B}{x^2}\) must be included in the decomposition.

System of Equations

After setting up the right form with the linear factors, we're faced with a system of equations to find the elusive coefficients A, B, and C. This system arises when equating the original expression to the decomposed form and multiplying through by the denominator to clear fractions.

Once the fractions are cleared, we obtain an equation where coefficients of like terms on both sides must match. For our example, after clearing fractions and reorganizing, the equation \(4x^2 + 2x - 1 = Ax(x + 1) + Bx + Cx^2\) results in a system that might look daunting. But fear not—it's just a matter of matching the coefficients of corresponding powers of \(x\) from both sides of the equation.

By setting up equations based on the coefficient comparison, we eventually solve for A, B, and C, with each variable usually pertaining to the coefficient of a corresponding power of \(x\). It's a little like detective work; finding the culprits—A, B, C—that when plugged back into the decomposed form, reconstruct the original expression perfectly.

Keep in Mind

Careful organization and solving the system accurately are essential for a correct decomposition, as these constants are the key to simplifying the original rational expression.