The given expression is:

$\frac{2x^2 - 4x}{2x^3 - 8x}$

Step-by-Step Simplification:

1. Factor both the numerator and the denominator:

   • Numerator: $2x^2 - 4x$ $= 2x(x - 2)$

   • Denominator: $2x^3 - 8x$ $= 2x(x^2 - 4) = 2x(x - 2)(x + 2)$

2. Rewrite the expression using these factored forms: $\frac{2x(x - 2)}{2x(x - 2)(x + 2)}$

3. Cancel out common factors: The common factors $2x(x - 2)$ appear in both the numerator and denominator, so they cancel out (as long as $x \neq 0$ and $x \neq 2$):

   $\frac{1}{x + 2}$

Final Answer:

$\frac{2x^2 - 4x}{2x^3 - 8x} = \frac{1}{x + 2} \quad \text{(for } x \neq 0, \, x \neq 2 \text{)}$

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Related Questions:

1. How do you factor polynomials like $2x^2 - 4x$?
2. What are the conditions under which we can cancel factors in fractions?
3. Can you explain how to factor differences of squares, like $x^2 - 4$?
4. What are some restrictions that arise when simplifying rational expressions?
5. How can you determine the domain of rational functions like $\frac{1}{x + 2}$?

Tip: Always check for restrictions when simplifying, especially when cancelling factors in a fraction.