If A+B=\frac{x^{2}-8}{x+2} and A-B=\frac{-x^{2}+2x+4}{x+2} then what is B equal to ?
- A. \frac{x^{2}-4}{x^{2}+4x+4}
- B. \frac{x^{2}-4}{x^{2}-4x+4}
- C. \frac{2x^{2}-7x+3}{2x-1} ✓
- D. \frac{2x^{2}+7x-3}{2x-1}
Correct Answer: C. \frac{2x^{2}-7x+3}{2x-1}
Explanation
Subtracting (A-B) from (A+B) gives 2B = \frac{(x^2-8) - (-x^2+2x+4)}{x+2} = \frac{2x^2-2x-12}{x+2} = \frac{2(x-3)(x+2)}{x+2} = 2(x-3). Thus, B = x-3. Option (c) simplifies to \frac{(2x-1)(x-3)}{2x-1} = x-3.
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