What is the simplified form of #(x^2-25)/(x-5)#?

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Answer 1 · 2015-09-13T13:33:53

#(x^2-25)/(x-5) = x+5# with exclusion #x != 5#

Use the difference of squares identity:

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

to find:

#(x^2-25)/(x-5) = (x^2-5^2)/(x-5) = ((x-5)(x+5))/(x-5)#

#= (x-5)/(x-5)*(x+5) = x+5#

with exclusion #x != 5#

Note that if #x = 5# then both #(x^2-25)# and #(x-5)# are #0#, so #(x^2-25)/(x-5) = 0/0# is undefined.