No. You have an infinite number of terms. That automatically makes it not a polynomial. To show this explicitly for this function, note that your function is just
[tex]f(x) = \sum_{n=2}^\infty \left(\frac{x}{10}\right)^n = \sum_{n=0}^\infty \left(\frac{x}{10}\right)^n - 1 -x[/tex]
which diverges when [itex]|x| \geq 10[/itex] and when |x|/10 < 1 converges to
[tex]f(x) = \frac{1}{1-\frac{x}{10}} - 1 - x[/tex]