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How would I reduce the following fraction? $$\frac{4A^2-B^2}{4A^2-4AB+B^2}$$ I am not sure how I would reduce it.

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Sometimes a good strategy is to set one of the variables to $1$; in this case, you might try $A=1$. When you do that, I’m sure you’ll know how to factor top and bottom. Use the same coefficients for the unmodified polynomials. –  Lubin Jun 8 '12 at 17:32

2 Answers 2

The numerator is a difference of squares, so it factors: $$4A^2 - B^2 = (2A)^2 - B^2= (2A-B)(2A+B).$$

The denominator is a perfect square: $$4A^2 - 4AB + B^2 = (2A)^2 - 2(2A)B + B^2 = (2A-B)^2.$$ Then you can cancel one factor: $$\frac{4A^2 - B^2}{4A^2 - 4AB+B^2} = \frac{(2A-B)(2A+B)}{(2A-B)^2} = \frac{2A+B}{2A-B}.$$

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Hint $\, $ For $\rm\ c = 2a\ $ it is $\rm\ \dfrac{c^2-b^2}{c^2-2cb-b^2},\ $ an obvious difference of squares over a perfect square.

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