To find the sum of the rational expressions \(\frac{x - 4}{-4} + \frac{2x - 8}{x^2 - 2x}\), we first need to simplify the second expression \(\frac{2x - 8}{x^2 - 2x}\).
- Factor the second expression:
- The numerator \(2x - 8\) can be factored to \(2(x - 4)\).
- The denominator \(x^2 - 2x\) can be factored to \(x(x - 2)\).
Thus, we can rewrite the second fraction: \[ \frac{2x - 8}{x^2 - 2x} = \frac{2(x - 4)}{x(x - 2)}. \]
Now, our sum becomes: \[ \frac{x - 4}{-4} + \frac{2(x - 4)}{x(x - 2)}. \]
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Find a common denominator: The common denominator for these two fractions is \(-4x(x - 2)\).
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Rewrite each fraction with the common denominator:
- The first fraction becomes: \[ \frac{x - 4}{-4} \cdot \frac{x(x - 2)}{x(x - 2)} = \frac{(x - 4)x(x - 2)}{-4} = \frac{-(x - 4)x(x - 2)}{4}. \]
- The second fraction becomes: \[ \frac{2(x - 4)}{x(x - 2)} \cdot \frac{-4}{-4} = \frac{-8(x - 4)}{4x(x - 2)}. \]
Now we can add the two fractions: \[ \frac{-(x - 4)x(x - 2) - 8(x - 4)}{4(-x(x - 2))}. \]
- Combine the numerators: Factor out \(-(x - 4)\) from the numerator: \[ -(x - 4) \left(x(x - 2) + 8\right) = -(x - 4)(x^2 - 2x + 8). \]
So our expression now becomes: \[ \frac{-(x - 4)(x^2 - 2x + 8)}{4(-x(x - 2))} = \frac{(x - 4)(x^2 - 2x + 8)}{4x(x - 2)}. \]
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Final simplification: We have the final expression as: \[ \frac{(x - 4)(x^2 - 2x + 8)}{4x(x - 2)}. \]
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Choose the correct option: Now, we can compare this expression with the options provided:
- A. \( \frac{2}{x + 2} \)
- B. \( \frac{2x}{x^4 - 2x} \)
- C. \( \frac{2x - 8}{x^2 - 2x - 8} \)
- D. \( \frac{x - 4}{x^2 - 4} \)
None of the options match our derived expression. It's possible that there may have been a mistake in the options or the problem setup.
However, the simplified sum of the given rational expressions is: \[ \frac{(x - 4)(x^2 - 2x + 8)}{4x(x - 2)}. \]
If you have further clarifications or need different details, feel free to ask!