the zero of the polynomial 2x ^2009+3x^2008+2x^2007+3x^2006+.......+2x+3 is:
Here's a slightly different approach:
The polynomial can be written as follows:
$\displaystyle p(x) = \sum\limits_{m = 0}^{1004} {x^{2m} } + \sum\limits_{n = 1}^{1005} {x^{2n - 1} } $
now from the rational root theorem we know that the rational roots of the polynomial must be from the following group of possible solutions:
$\displaystyle \pm 1, \pm \frac{1}
{2}, \pm \frac{3}
{2}, \pm 3$
the positive solutions are obviously impossible, so now using the formula for the sum of the geometric progression we can easily find that the correct solution is:
$\displaystyle
- \frac{3}
{2}
$