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The given expression to find the partial fraction decomposition is:

\[
\frac{x^2 + 3x - 8}{(x+1)(x^2 - 2)}
\]

### Step 1: Set up the partial fractions
The given denominator \((x+1)(x^2 - 2)\) suggests that the partial fraction decomposition will be of the form:

\[
\frac{x^2 + 3x - 8}{(x+1)(x^2 - 2)} = \frac{A}{x+1} + \frac{Bx + C}{x^2 - 2}
\]

### Step 2: Multiply by the denominator
Multiply both sides by \((x+1)(x^2 - 2)\) to eliminate the denominator:

\[
x^2 + 3x - 8 = A(x^2 - 2) + (Bx + C)(x + 1)
\]

### Step 3: Expand and combine like terms
Expand both sides:

\[
x^2 + 3x - 8 = A(x^2 - 2) + (Bx + C)(x + 1)
\]

\[
x^2 + 3x - 8 = A(x^2 - 2) + Bx^2 + Bx + Cx + C
\]

\[
x^2 + 3x - 8 = (A + B)x^2 + (B + C)x + (C - 2A)
\]

### Step 4: Equate coefficients
Now, equate the coefficients of corresponding powers of \(x\) on both sides:

For \(x^2\) term:
\[
A + B = 1
\]

For \(x\) term:
\[
B + C = 3
\]

For constant term:
\[
C - 2A = -8
\]

### Step 5: Solve the system of equations
We now solve the system of equations:

1. \(A + B = 1\)
2. \(B + C = 3\)
3. \(C - 2A = -8\)

From the first equation, we get:
\[
B = 1 - A
\]

Substitute \(B = 1 - A\) into the second equation:
\[
(1 - A) + C = 3
\]
\[
C = 2 + A
\]

Substitute \(C = 2 + A\) into the third equation:
\[
(2 + A) - 2A = -8
\]
\[
2 - A = -8
\]
\[
A = 10
\]

Now substitute \(A = 10\) back into the first and second equations:
\[
B = 1 - 10 = -9
\]
\[
C = 2 + 10 = 12
\]

### Step 6: Write the final partial fraction decomposition
Substituting \(A = 10\), \(B = -9\), and \(C = 12\) into the partial fraction decomposition:

\[
\frac{x^2 + 3x - 8}{(x+1)(x^2 - 2)} = \frac{10}{x+1} + \frac{-9x + 12}{x^2 - 2}
\]

This is the partial fraction decomposition of the given function.

Do you want details on any step or have further questions?

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**Related Questions:**
1. How would you handle partial fractions when the denominator contains repeated irreducible quadratic factors?
2. What is the importance of equating coefficients in partial fraction decomposition?
3. Can partial fraction decomposition be used for solving integrals? How?
4. How does partial fraction decomposition differ when the numerator is of higher degree than the denominator?
5. How do you perform partial fraction decomposition when the denominator has complex roots?

**Tip:** Always check if the numerator can be factored or simplified before performing partial fraction decomposition; this can sometimes simplify the process.

What is the partial fraction decomposition of the given function?

Learn how to perform partial fraction decomposition on a rational function with detailed steps. This guide covers algebraic techniques including equating coefficients, suitable for students in Grades 10-12.

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