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Maybe I am just too tired, but I dont know how to solve this??

Can you point me in right direction. Thanks!!!

depecheSoul
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    As an odd polynomial it has at least (and in fact only) one real root. However, don't see any easy way other than numerical methods to find it. – Macavity Mar 09 '14 at 15:27

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There is one real root, approximately $-0.59622$; you can find it using the bisection method. It's not "nice", unfortunately.

The bisection method:

Start with real numbers $L,R$, such that $f(L), f(R)$ are of different signs, and $L<R$. Now, set $T$ (for trial) to be $T=\frac{L+R}{2}$. Evaluate $f(T)$. If it is the same sign as $L$, then start over using $T,R$ as the new interval. If instead $f(T)$ is the same sign as $R$, then start over using $L,T$ as the new interval.

This method cuts the interval's length in half every time, so it converges quite quickly -- one binary digit every step.

vadim123
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As said by vadim123, there is only one real root. By inspection, you can locate it between $x=-1$ and $x=0$. At this point, you can use the classical bisection method or, if you want a faster convergence, Newton method.

Just be lazy (as I am) and start iterating at $x=-1$. The successive iterates will then be $-0.794118$, $-0.657914$, $-0.603276$, $-0.596322$, $-0.596225$ which is the solution.

If you start iterating at $x=0$, the first iterate will be $-1.0000$ and we should be back to the previous scheme.

Claude Leibovici
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