Re: 2012 HSC MX2 Marathon
Let's see if this works...
p(x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + ... + \frac{x^{2n}}{2n!}
We know that n is an integer, so 2n must be even. Therefore, p(x) is an even polynomial, as it has a degree of 2n.
From the fundamental theorem of algebra, we also...