Tricky Factorization How do I factor this expression:  $$ 0.09e^{2t} + 0.24e^{-t} + 0.34 + 0.24e^t + 0.09e^{-2t} ? $$ 
By trial and error I got $$ \left(0.3e^t + 0.4 + 0.3 e^{-t}\right)^2$$ but I'd like to know how to formally arrive at it.
Thanks.

The most striking thing about the given expression is  the symmetry.  For anything with that kind of symmetric structure, there is a systematic approach which is definitely not trial and error.
Let 
$$z=e^t+e^{-t}.$$
Square. We obtain
$$z^2=e^{2t}+2+e^{-2t},$$
and therefore $e^{2t}+e^{-2t}=z^2-2$.
Substitute in our expression. We get
$$0.09(z^2-2)+0.24 z+0.34.\qquad\qquad(\ast)$$
This simplifies to
$$0.09z^2 +0.24z +0.16.$$
The factorization is now obvious. We recognize that we have simply $(0.3z+0.4)^2$. Now replace $z$ by $e^t+e^{-t}$.
If the numbers had been a little different (but still with the basic $e^t$, $e^{-t}$ symmetry) we would at the stage $(\ast)$ obtain some other quadratic.  In general, quadratics with real roots can be factored as a product of linear terms. It is just a matter of completing the square. For example, replace the constant term $0.34$ by, say, $0.5$. We get a quadratic in $z$ that does not factor as prettily, but it does factor. 
Comment: For fun we could instead make the closely related substitution $2y=e^t+e^{-t}$, that is, $y=\cosh t$.  If we analyze the substitution process further, we get to useful pieces of mathematics, such as the Chebyshev polynomials.
The same idea is the standard approach  to finding the roots of palindromic polynomials. For example, suppose that we want to solve the equation $x^4 +3x^3-10x^2+3x+1=0$.  Divide through by $x^2$. We get the equation
$$x^2+3x-10+\frac{3}{x}+\frac{1}{x^2}=0.$$
Make the substitution $z=x+\frac{1}{x}$. Then $x^2+\frac{1}{x^2}=z^2-2$. Substitute. We get a quadratic in $z$.
