How to simplify this .?

#(4x^2 - y^2)/(12x^2 - 4xy-y^2)#

1 Answer
George C.
Jul 31, 2017

#(4x^2-y^2)/(12x^2-4xy-y^2) = (2x+y)/(6x+y)#

with exclusion #2x != y#

Explanation:

Given:

#(4x^2-y^2)/(12x^2-4xy-y^2)#

We can factor both numerator and denominator and then cancel any common factor.

The numerator is a difference of squares, so factors as:

#4x^2-y^2 = (2x)^2-y^2 = (2x-y)(2x+y)#

To factor the denominator find a pair of factors of #12# which differ by #4#. The pair #6, 2# works, so use that to split the middle term and factor by grouping:

#12x^2-4xy-y^2 = (12x^2-6xy)+(2xy-y^2)#

#color(white)(12x^2-4xy-y^2) = 6x(2x-y)+y(2x-y)#

#color(white)(12x^2-4xy-y^2) = (6x+y)(2x-y)#

So we find:

#(4x^2-y^2)/(12x^2-4xy-y^2) = (color(red)(cancel(color(black)((2x-y))))(2x+y))/((6x+y)color(red)(cancel(color(black)((2x-y)))))#

#color(white)((4x^2-y^2)/(12x^2-4xy-y^2)) = (2x+y)/(6x+y)#

with exclusion #2x != y#

Related questions
See all questions in Multiplication and Division of Radicals
Impact of this question
933 views around the world
You can reuse this answer
Creative Commons License