Problem 22 - College Algebra Essentials

Question

write the partial fraction decomposition of each rational expression. $$ \frac{x}{(x+1)^{2}} $$

Step-by-Step Solution

Answer

The partial fraction decomposition of the given expression is \(\frac{0}{x+1} + \frac{0}{(x+1)^2}\), which simplifies to 0.

Step 1: Express the fraction in general form of partial fractions

Since we have a repeated factor \((x+1)^2\) in the denominator, the general form of the partial fractions would be \[\frac{x}{(x+1)^2} = \frac{A}{x+1} + \frac{B}{(x+1)^2}\] where A and B are unknown constants that we need to determine.

Step 2: Clear the denominator

This can be done by multiplying both sides of the equation by \((x+1)^2\), which results in a new equation \(x = A(x+1) + B\). This equation works for all x except x=-1.

Step 3: Solve for coefficients A and B

Choose convenient values for x to make the equation simpler. We can start by choosing x=-1. This gives us 0 = A(0) + B which simplifies to B=0. Next, we choose x=0 which gives 0 = A(1) + 0. Thus, A=0.

Key Concepts

Rational Expressions

A rational expression is essentially a fraction that has a polynomial in both its numerator and its denominator. The beauty of rational expressions is that they can be simplified, decomposed, or manipulated just like regular numbers but with polynomials. In this exercise, the rational expression is given by \( \frac{x}{(x+1)^{2}} \). Here, \(x\) is the numerator, while \((x+1)^2\) is the denominator.

Understanding rational expressions is key to moving forward with partial fraction decomposition. The goal of decomposing rational expressions into partial fractions is to express a complex fraction as a sum of simpler fractions. This can make integration and other algebraic processes more manageable. In our problem, this means breaking down \( \frac{x}{(x+1)^2} \) into simpler terms that are easier to work with.

Repeated Factors

Repeated factors in the denominator of a rational expression require special consideration during partial fraction decomposition. When a factor is repeated, such as \((x+1)^2\), it means that the factor \((x+1)\) appears twice in the denominator. To account for each occurrence, each power of the repeated factor gets its own term in the decomposition.

For example, in the expression \( \frac{x}{(x+1)^2} \), since \( (x+1) \) is repeated, the partial fraction decomposition will include terms for both \( \frac{A}{x+1} \) and \( \frac{B}{(x+1)^2} \).

• First, \( \frac{A}{x+1} \) represents the first occurrence of the factor.
• Second, \( \frac{B}{(x+1)^2} \) covers the second occurrence, accounting for the square.

The inclusion of terms for each power of a repeated factor ensures the correct distribution of values across the expression, allowing you to solve for the unknown constants later.

Unknown Constants

When decomposing a rational expression into partial fractions, unknown constants like \(A\) and \(B\) play a critical role. They are placeholders that, once found, complete the decomposition. Determining these constants involves creating equations based on equality, as seen in our exercise.

After setting the initial equation \( \frac{x}{(x+1)^2} = \frac{A}{x+1} + \frac{B}{(x+1)^2} \), we eliminate the denominators by multiplying through by \((x+1)^2\). This yields \( x = A(x+1) + B \). The next step is to solve for \(A\) and \(B\). A convenient way to do this is to substitute strategic values of \(x\):

• Choosing \( x = -1 \) simplifies \( A(0) + B = 0 \), giving \( B = 0 \).
• Choosing another value, like \( x = 0 \), simplifies to \( 0 = A(1) + 0 \), resulting in \( A = 0 \).

By selecting values for \(x\) that make terms drop out, it's easier to solve for the unknown constants effectively.