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**Find the partial fraction decomposition of:**

\[
\frac{5x^2}{(x - 2)^2 (x^2 + 2^2)}
\]

### Step-by-Step Solution

For partial fraction decomposition, the rational function will be decomposed into simpler terms that correspond to the factors in the denominator. The factors in the denominator are \((x - 2)^2\) and \((x^2 + 4)\).

### General Form for Partial Fraction Decomposition

1. For the \((x - 2)^2\) factor, since it is a repeated factor, the decomposition includes:
   - \(\frac{A}{x - 2} + \frac{B}{(x - 2)^2}\)

2. For the quadratic factor \((x^2 + 4)\), since it's an irreducible quadratic, we use:
   - \(\frac{Cx + D}{x^2 + 4}\)

So the decomposition will have the form:

\[
\frac{5x^2}{(x - 2)^2 (x^2 + 4)} = \frac{A}{x - 2} + \frac{B}{(x - 2)^2} + \frac{Cx + D}{x^2 + 4}
\]

### Step 1: Multiply by the common denominator

To eliminate the denominators, multiply both sides of the equation by \((x - 2)^2(x^2 + 4)\):

\[
5x^2 = A(x - 2)(x^2 + 4) + B(x^2 + 4) + (Cx + D)(x - 2)^2
\]

### Step 2: Expand each term

1. Expand \(A(x - 2)(x^2 + 4)\):
   \[
   A(x - 2)(x^2 + 4) = A(x^3 - 2x^2 + 4x - 8)
   \]
   
2. Expand \(B(x^2 + 4)\):
   \[
   B(x^2 + 4) = Bx^2 + 4B
   \]

3. Expand \((Cx + D)(x - 2)^2\):
   \[
   (x - 2)^2 = x^2 - 4x + 4
   \]
   \[
   (Cx + D)(x - 2)^2 = (Cx + D)(x^2 - 4x + 4) = Cx^3 - 4Cx^2 + 4Cx + Dx^2 - 4Dx + 4D
   \]
   \[
   = Cx^3 + (-4C + D)x^2 + (4C - 4D)x + 4D
   \]

### Step 3: Combine all terms

Now, add up all the expanded terms:

\[
5x^2 = A(x^3 - 2x^2 + 4x - 8) + (Bx^2 + 4B) + (Cx^3 + (-4C + D)x^2 + (4C - 4D)x + 4D)
\]

Combine powers of \(x\):

\[
5x^2 = (A + C)x^3 + (-2A + B - 4C + D)x^2 + (4A + 4C - 4D)x + (-8A + 4B + 4D)
\]

### Step 4: Set up a system of equations

Since this equation must be true for all \(x\), the coefficients of like powers of \(x\) on both sides must be equal. This gives the following system of equations:

1. Coefficient of \(x^3\): \(A + C = 0\)
2. Coefficient of \(x^2\): \(-2A + B - 4C + D = 5\)
3. Coefficient of \(x^1\): \(4A + 4C - 4D = 0\)
4. Constant term: \(-8A + 4B + 4D = 0\)

### Step 5: Solve the system of equations

From equation (1): \(A + C = 0\) ⇒ \(A = -C\).

Substitute \(A = -C\) into the remaining equations:

2. \(-2(-C) + B - 4C + D = 5\) ⇒ \(2C + B - 4C + D = 5\) ⇒ \(-2C + B + D = 5\)
3. \(4(-C) + 4C - 4D = 0\) ⇒ \(-4C + 4C - 4D = 0\) ⇒ \(-4D = 0\) ⇒ \(D = 0\)
4. \(-8(-C) + 4B + 4(0) = 0\) ⇒ \(8C + 4B = 0\) ⇒ \(4B = -8C\) ⇒ \(B = -2C\)

### Step 6: Substitute into the equations

Now substitute \(B = -2C\) and \(D = 0\) into the equation \(-2C + B + D = 5\):

\[
-2C - 2C + 0 = 5 \quad \Rightarrow \quad -4C = 5 \quad \Rightarrow \quad C = -\frac{5}{4}
\]

So:
- \(A = -C = \frac{5}{4}\)
- \(B = -2C = \frac{5}{2}\)
- \(D = 0\)

### Final Answer:

The partial fraction decomposition is:

\[
\frac{5x^2}{(x - 2)^2(x^2 + 4)} = \frac{5/4}{x - 2} + \frac{5/2}{(x - 2)^2} + \frac{-5/4x}{x^2 + 4}
\]

Would you like further clarification or additional details? Here are five related questions you might find helpful:

1. How do you solve systems of equations with fractions?
2. What is the process for decomposing rational expressions with repeated factors?
3. Why do we use constants in the form \(Cx + D\) for irreducible quadratics?
4. How do you expand and simplify polynomial expressions in partial fractions?
5. How can partial fraction decomposition be applied to integration?

**Tip:** For solving partial fractions, always pay close attention to repeated and irreducible factors—each requires a specific form in the decomposition.

Write the partial fraction decomposition of the given rational expression: 5x^2 / ((x-2)^2 (x^2 + 4))

Learn how to find the partial fraction decomposition of 5x^2 / ((x - 2)^2(x^2 + 4)) step by step. This solution covers rational expressions, factoring, and solving systems of equations for partial fraction coefficients. Ideal for precalculus students.

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