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How do you simplify and state the excluded values for #(x+5) /(x^2-25)#?

1 Answer

May 26, 2015

Note that #(x+5)/(x^2-25)# is only defined if the denominator is not equal to zero.
That is #x^2-25 != 0#
#rarr x=5# and #x=-5# are excluded values.

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

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

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

#= 1/(x-5)#

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