How do you simplify #(4w^2-49z^2)/(14z-4w)#?

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Answer 1 · 2018-06-11T14:07:43

#=-(2w+7z)/2#

Factorise both expressions

#(4w^2-49z^2)/(14z-4w)" "(larr"difference of squares")/(" "larr "common factor")#

#=((2w+7z)(2w-7z))/(2(7z-2w))#

take out a common factor of #-1# from the denominator

#=((2w-7z)(2w+7z))/(-2(-7z+2w)) " "rarr=((2w-7z)(2w+7z))/(-2(2w-7z))#

# (cancel((2w-7z))(2w+7z))/(-2cancel((2w-7z)))#

#=-(2w+7z)/2#

Answer 2 · 2018-06-12T13:53:17

#color(blue)(=> -(2w + 7z) / 2#

#(4w^2 - 49z^2) / (14z - 4w)#

Numerator is in the form #(a^2 - b^2) = (a+b)(a-b)#

#=> ((2w + 7z) (2w - 7z)) / (2 * (7z - 2w))#

#=> -((2w + 7z) * cancel (7z - 2w)) / (2 * cancel(7z - 2w))#

#=>- (2w + 7z) / 2#