How do you find the quotient of #(4x^2 – y^2) ÷ (2x – y)#?

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Answer 1 · 2015-10-29T04:34:42

#2x +y #

Note that

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

Since #a^2- b^2 = (a+b) (a-b)#

Answer 2 · 2015-10-29T04:37:59

(2x +y)

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

( 4#x^2# - #y^2#) can be written as #( 2x +y) (2x -y)#

#(2x +y ) ( 2x-y) // (2x-y) #

#2x +y#