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 $(a^2 - b^2) = (a+b)(a-b)$

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

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

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

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