Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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

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: $$\frac{4A^2 - B^2}{4A^2 - 4AB+B^2} = \frac{(2A-B)(2A+B)}{(2A-B)^2} = \frac{2A+B}{2A-B}.$$

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.