How do you simplify #(x^2-4x)/(16-x^2)#?

1 Answer
Alan N.
Aug 15, 2016

#-x/(x+4)# (If #x!=4#)

Explanation:

#(x^2-4x)/(16-x^2) = (x(x-4))/(-(x^2-16))#

Given: #(a^2-b^2) = (a+b)(a-b)#

#(x(x-4))/(-(x^2-16)) =- (x(x-4))/((x+4)(x-4))#

#=- (xcancel((x-4)))/((x+4)cancel((x-4)))# (If #x!=4#)

#=-x/(x+4)#

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