# Question #43bd9

Oct 17, 2017

${\left({e}^{x} + {e}^{-} x\right)}^{2}$

#### Explanation:

I'm assuming you mean ${e}^{2 x} + 2 + {e}^{- 2 x}$.

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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:

${a}^{b c} = {\left({a}^{b}\right)}^{c}$

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

${e}^{2 x} + 2 + {e}^{- 2 x}$

${\left({e}^{x}\right)}^{2} + 2 + {\left({e}^{-} x\right)}^{2}$

Also, remember that ${e}^{-} x$ is the same as $\frac{1}{e} ^ x$.

${\left({e}^{x}\right)}^{2} + 2 + {\left(\frac{1}{e} ^ x\right)}^{2}$

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

${\left(a + b\right)}^{2} = {a}^{2} + 2 a b + {b}^{2}$

Well, if we use the fact that $\left({e}^{x}\right) \left(\frac{1}{e} ^ x\right) = 1$, we can change our polynomial to look like this:

${\left({e}^{x}\right)}^{2} + 2 \left({e}^{x}\right) \left(\frac{1}{e} ^ x\right) + {\left(\frac{1}{e} ^ x\right)}^{2}$

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

${\left({e}^{x} + \frac{1}{e} ^ x\right)}^{2}$

Or, to write it like the original problem did,

${\left({e}^{x} + {e}^{-} x\right)}^{2}$