How do you simplify #(x^2-8x+16 )/(x^2-12x+32 )#?

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Answer 1 · 2016-04-08T21:33:46

#(x-4)/(x-8)#

Consider#" "x^2-8x+16#

Notice that #4xx4=16" and "4+4=8#

#(x-4)(x-4)=(x-4)^2#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Consider #" "x^2-12x+32#

notice that #4xx8=32" and "8+4=12#

#(x-4)(x-8)#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Putting it all together")#

#(x^2-8x+16)/(x^2-12x+32) = ((x-4)(x-4))/((x-4)(x-8))#

#=>(x-4)/(x-4)xx(x-4)/(x-8)#

But #(x-4)/(x-4) =1# giving:

#(x-4)/(x-8)#