Simplify and state any excluded values? C-5/C^2-25

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Answer 1 · 2018-07-29T15:23:17

#color(purple)(=>1/(c + 5)# for #color(crimson)(c != 5#

#(c - 5) / (c^2 - 25)#

#=> cancel(c - 5)^color(red)(1) / ((c + 5) cancel(c - 5))#

#=> 1 / (c + 5)# for #c != 5#

As when #c = 5#, Numerator and Denominator become #0#.

Hence the term becomes #=>0/0# which is indeterminant.

Answer 2 · 2018-07-29T15:50:11

#1/(c+5), c!=5#

We have the following:

#(c-5)/(c^2-25)#

Notice that the denominator is a difference of squares

#a^2-b^2#, that can be factored as #(a+b)(a-b)#. This simplifies to

#color(blue)((c-5)/((c+5)(c-5)))#

We can cancel out common factors on the top and bottom to get

#cancel(c-5)/((c+5)cancel(c-5))=1/(c+5)#

Recall that in the initial expression (blue), if #c=5#, that would make this expression undefined, as we would get indeterminate form

#0/0#

Because of this, in our simplified expression, #c# still cannot be #5#.

#1/(c+5), c!=5#

Hope this helps!