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December 20, 2014

December 20, 2014

Posted by **Praphul** on Saturday, January 5, 2013 at 2:09pm.

- Math -
**Steve**, Saturday, January 5, 2013 at 3:45pmthere are 5 possible rational and real roots. Given the wording of the problem, though, it's likely that there are some complex roots, reducing the number of real candidates.

There may be 2 or 4 complex roots, meaning that there may be 5,3, or 1 real root.

Lacking any further information, all the real roots may be rational, and in fact will be integers.

A little synthetic division shows that we have no rational roots.

In fact, there are no real roots greater than 0 or less than -1. So, there are 4 complex roots and one real root -1 < r < 0

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