Question #43bd9

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Question #43bd9

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Answer 1 · 2017-10-17T22:25:11

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

Explanation:

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

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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:

$a^{b c} = {(a^{b})}^{c}$ $a^(bc) = (a^b)^c$

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

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

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

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

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

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

${(a + b)}^{2} = a^{2} + 2 a b + b^{2}$ $(a+b)^2 = a^2 + 2ab + b^2$

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

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

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

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

Or, to write it like the original problem did,

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

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Question #43bd9

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