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To express the function \(\frac{2x - 8}{x^2 - 6x + 5}\) as a power series centered at \(x = 0\), we need to follow these steps:

### Step 1: Partial Fraction Decomposition

The first step is to decompose the function into partial fractions. We start by factoring the denominator:

\[ 
x^2 - 6x + 5 = (x - 1)(x - 5).
\]

Thus, we can write:

\[
\frac{2x - 8}{x^2 - 6x + 5} = \frac{2x - 8}{(x - 1)(x - 5)} = \frac{A}{x - 1} + \frac{B}{x - 5}.
\]

To find \(A\) and \(B\), we solve the equation:

\[
2x - 8 = A(x - 5) + B(x - 1).
\]

Expanding and simplifying:

\[
2x - 8 = Ax - 5A + Bx - B = (A + B)x - (5A + B).
\]

Equating coefficients, we get:

1. \(A + B = 2\)
2. \(-5A - B = -8\)

Solving these equations simultaneously:

1. From \(A + B = 2\), we have \(B = 2 - A\).
2. Substitute into the second equation: \(-5A - (2 - A) = -8\).

Simplify:

\[
-5A - 2 + A = -8 \quad \Rightarrow \quad -4A = -6 \quad \Rightarrow \quad A = \frac{3}{2}.
\]

Then, \(B = 2 - \frac{3}{2} = \frac{1}{2}\).

So the partial fraction decomposition is:

\[
\frac{2x - 8}{(x - 1)(x - 5)} = \frac{\frac{3}{2}}{x - 1} + \frac{\frac{1}{2}}{x - 5}.
\]

### Step 2: Express as Power Series

Now, express each term as a power series centered at \(x = 0\):

1. For \(\frac{\frac{3}{2}}{x - 1}\), rewrite it as \(-\frac{3}{2} \cdot \frac{1}{1 - x}\). The geometric series expansion is:

   \[
   \sum_{n=0}^{\infty} (-\frac{3}{2})x^n \quad \text{for} \quad |x| < 1.
   \]

2. For \(\frac{\frac{1}{2}}{x - 5}\), rewrite it as \(-\frac{1}{10} \cdot \frac{1}{1 - \frac{x}{5}}\). The geometric series expansion is:

   \[
   \sum_{n=0}^{\infty} \left(-\frac{1}{10}\right) \left(\frac{x}{5}\right)^n = \sum_{n=0}^{\infty} \left(-\frac{1}{10}\right) \frac{x^n}{5^n} \quad \text{for} \quad |x/5| < 1 \Rightarrow |x| < 5.
   \]

### Step 3: Power Series Representation

Combining both series:

\[
E(x) = \sum_{n=0}^{\infty} \left[(-\frac{3}{2})x^n + \left(-\frac{1}{10}\right) \frac{x^n}{5^n}\right].
\]

Simplifying the series:

\[
E(x) = \sum_{n=0}^{\infty} \left(-\frac{3}{2} - \frac{1}{10 \times 5^n}\right)x^n.
\]

### Step 4: Interval of Convergence

The interval of convergence is determined by the stricter condition of the two series:

1. \(|x| < 1\) for \(\frac{\frac{3}{2}}{x - 1}\).
2. \(|x| < 5\) for \(\frac{\frac{1}{2}}{x - 5}\).

Therefore, the interval of convergence for the combined series is \(|x| < 1\).

### Final Answer

- **Power Series Representation:**
  \[
  E(x) = \sum_{n=0}^{\infty} \left(-\frac{3}{2} - \frac{1}{10 \times 5^n}\right)x^n
  \]

- **Interval of Convergence:** \((-1, 1)\)

If you have any questions or need further details, feel free to ask!

Here are some additional questions to explore:

1. How is partial fraction decomposition helpful in finding power series?
2. What are the properties of geometric series?
3. How does the interval of convergence relate to the radius of convergence?
4. Can partial fraction decomposition be used for higher degree polynomials?
5. What other methods exist for finding power series representations?
6. How do complex numbers affect series expansions?
7. What are the implications of different intervals of convergence?
8. How can you verify the correctness of a series expansion?

**Tip:** Remember that the geometric series formula \(\sum_{n=0}^{\infty} ar^n = \frac{a}{1-r}\) is crucial for converting fractions into power series!

Explore how to express \(\frac{2x - 8}{x^2 - 6x + 5}\) as a power series centered at \(x = 0\). Learn step-by-step partial fraction decomposition and geometric series expansions with detailed examples. Suitable for advanced mathematics students.

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