Question #43bd9

1 Answer
Oct 17, 2017

Answer:

#(e^x + e^-x)^2#

Explanation:

I'm assuming you mean #e^(2x) + 2 + e^(-2x)#.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

There's something very special about this polynomial. To see it, let's use one of the rules of exponents to write the first and last terms in a different way:

#a^(bc) = (a^b)^c#

Therefore, we can change our polynomial to look like this:

#e^(2x) + 2 + e^(-2x)#

#(e^x)^2 + 2 + (e^-x)^2#

Also, remember that #e^-x# is the same as #1/e^x#.

#(e^x)^2 + 2 + (1/e^x)^2#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Do you see it? This polynomial is actually a perfect square! Remember this formula?

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

Well, if we use the fact that #(e^x)(1/e^x) = 1#, we can change our polynomial to look like this:

#(e^x)^2 + 2(e^x)(1/e^x) + (1/e^x)^2#

This very clearly fits with our perfect square formula, so we can factor it like this:

#(e^x + 1/e^x)^2#

Or, to write it like the original problem did,

#(e^x + e^-x)^2#

Final Answer