Resolve this into partial fractions: (x^3 - 2)/(((x + 1)^2)(x-1))

To decompose the rational function \(\frac{x^3 - 2}{(x + 1)^2(x - 1)}\) into partial fractions, we follow these steps:

### Step 1: Set up the form of the partial fraction decomposition

Since the denominator \((x + 1)^2(x - 1)\) consists of linear factors (some repeated), the partial fraction decomposition should have the following form:

\[
\frac{x^3 - 2}{(x + 1)^2(x - 1)} = \frac{A}{x + 1} + \frac{B}{(x + 1)^2} + \frac{C}{x - 1}
\]

### Step 2: Multiply through by the denominator

To eliminate the fractions, multiply both sides by \((x + 1)^2(x - 1)\):

\[
x^3 - 2 = A(x + 1)(x - 1) + B(x - 1) + C(x + 1)^2
\]

### Step 3: Expand and collect like terms

Expand each term on the right-hand side:

1. \(A(x + 1)(x - 1)\) simplifies to \(A(x^2 - 1)\).
2. \(B(x - 1)\) stays as \(B(x - 1)\).
3. \(C(x + 1)^2\) expands to \(C(x^2 + 2x + 1)\).

So, we have:

\[
x^3 - 2 = A(x^2 - 1) + B(x - 1) + C(x^2 + 2x + 1)
\]

Now expand and collect the terms:

\[
x^3 - 2 = (A + C)x^2 + (B + 2C)x + (-A - B + C)
\]

### Step 4: Equate coefficients

Now, compare the coefficients of \(x^3\), \(x^2\), \(x\), and the constant term on both sides of the equation:

- For \(x^2\): \(A + C = 0\)
- For \(x\): \(B + 2C = 0\)
- For the constant term: \(-A - B + C = -2\)

### Step 5: Solve the system of equations

Now, solve the system of linear equations:

1. \(A + C = 0 \implies A = -C\)
2. \(B + 2C = 0 \implies B = -2C\)
3. \(-A - B + C = -2\)

Substitute \(A = -C\) and \(B = -2C\) into the third equation:

\[
-(-C) - (-2C) + C = -2
\]

Simplify:

\[
C + 2C + C = -2 \implies 4C = -2 \implies C = -\frac{1}{2}
\]

Now substitute \(C = -\frac{1}{2}\) into the other equations:

\[
A = -C = \frac{1}{2}, \quad B = -2C = 1
\]

### Step 6: Write the final partial fraction decomposition

Substitute \(A\), \(B\), and \(C\) back into the partial fraction form:

\[
\frac{x^3 - 2}{(x + 1)^2(x - 1)} = \frac{\frac{1}{2}}{x + 1} + \frac{1}{(x + 1)^2} + \frac{-\frac{1}{2}}{x - 1}
\]

This simplifies to:

\[
\frac{x^3 - 2}{(x + 1)^2(x - 1)} = \frac{1}{2(x + 1)} + \frac{1}{(x + 1)^2} - \frac{1}{2(x - 1)}
\]

This is the partial fraction decomposition.

Would you like more details, or do you have any questions?

---

Here are 5 related questions you might find interesting:

1. How would you decompose a fraction where the numerator has a higher degree than the denominator?
2. What happens if one of the roots in the denominator is a repeated quadratic factor?
3. Can partial fractions be used for integrals, and if so, how?
4. How would you handle a case with complex roots in the denominator?
5. How can you verify the correctness of the partial fraction decomposition?

**Tip:** Partial fraction decomposition is often useful in integration, particularly for rational functions where direct integration isn't straightforward.

Learn how to decompose the rational function (x^3 - 2)/((x + 1)^2(x - 1)) into partial fractions with a step-by-step solution. This problem covers key algebraic concepts, including partial fractions, rational functions, and solving systems of linear equations. Suitable for students in Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation