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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.

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up vote 17 down vote accepted

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 $z$.

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HINT $\ $ Exploit the innate symmetry in the expression using the following with $\rm\: X = e^t\:$

$$\rm\ A\ (X + X^{-1})^2 + B\ (X + X^{-1}) + C\ \ =\ \ A\ X^2 + B\ X + (2\:A + C) + B\ X^{-1} + A\ X^{-2}$$

In your example the LHS quadratic is a perfect square since it has discriminant $ = 0$, namely

$$\rm\ a^2\ (X + X^{-1})^2 + 2\:a\:b\ (X + X^{-1}) + b^2\ =\ (a\ (X + X^{-1}) + b)^2\quad\ for\ \ a = 0.3,\ b = 0.4 $$

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First of all, factoring is basically trial and error. In some cases, it can be done without guess and check. But, when you learn to factor polynomials, that is how you are taught to do it. You have a few ideas that give you the basic form, then you guess and check to see if it multiplies out right. The more you do it, the better you get at it. In cases like a quadratic polynomial, you can actually complete the square and find the answer, but for more complicated polynomials that no longer works.

So, if you want to factor this, there's got to be some guessing going on. But, you can make some educated guesses. You look at it and you see the highest power term of e is 2t and the lowest is -2t. You guess that this the square of something with an $e^t$ and an $e^{-t}$. You deduce the coefficients of these since the product has $0.09e^{2t}$ and $0.09e^{-2t}$. All that's really left is a constant term, $c$. Thus, you have $$(0.3 e^t + c + 0.3 e^{-t})^2$$ You multiply it out. You figure out that $c = 0.4$. Assuming this is the right basic form, the only possible choice for $c$ is $0.4$ since $0.24 e^t$ comes from $2c \cdot 0.3 e^t$.

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The first step is to get rid of the exponential by putting $x = e^t$: $$f(x) = 0.09x^2 + 0.24x + 0.34 + 0.24x^{-1} + 0.09x^{-2}.$$ The second step is to get rid of the negative powers by multiplying by $x^2$: $$g(x) = x^2f(x) = 0.09x^4 + 0.24x^3 + 0.34x^2 + 0.24x + 0.09.$$ Now we have a polynomial $g(x)$ that we need to factor. We get $$ g(x) = (0.3x^2 + 0.4x + 0.3)^2. $$ This implies a factorization of $f(x)$: $$ f(x) = x^{-2}f(x) = (0.3x + 0.4 + 0.3x^{-1})^2. $$ All you need to do now is to substitute $x = e^t$.

The advantage of this method is that it reduces the problem to factorization of polynomials, which is something we already know how to do.

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Writing $x = e^t$ makes the factorization more familiar, but it actually doesn't answer the OP's question. How did you know how to factor $g(x)$ as $g(x) = (0.3x^2 +0.4x + 0.3)$? – JavaMan Dec 16 '11 at 16:14
There are known algorithms for factoring polynomials over the rationals. They are probably already programmed in Wolfram alpha, so you can just plug it there and see what you get. No tricks here. – Yuval Filmus Dec 17 '11 at 0:09

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