I have eddited the Question, I hope the Statement is now true.
I want to show that every such polynomial has a zeropoint, If I can prove the Statement above, I can use the indermediate value Theorem.
What I have worked out so far is that
$P(x)$, can be rewritten in the form of:
$P(x)= a_0x^{2k+1}+a_1x^{i_1}+a_2x^{i_2}+...a_jx^{i_j}+C,j\in\mathbb{N}_0$ and $2k+1>i_1>…>i_j$
$= (a_0x^{2k+1-i_j}+a_1x^{i_1-i_j}+...+a_j)x^{i_j}+C$
$= ((a_0x^{2k+1-i_j-(i_{j-1}-i_j)}+...a_{j-1})x^{i_{j-1}-i_j}+a_j)x^{i_j}+C$
$=((a_0x^{2k+1-i_{j-1}}+...a_{j-1})x^{i_{j-1}-i_j}+a_j)x^{i_j}+C$
$...$
$=(…(a_0x^{2k+1-i_1}+a_1)x^{i_2-i_1}+a_2)…)x^{i_{j-1}-i_j}+a_j)x^{i_j}+C$
How can I continue, do you have any suggestions?
I had the idea to pick
$\max\limits_{0 \leq s \leq j}|a_s|$ as my $x$
I somehow think that $P(x)$ would either be positive or negative dependant whether I pick $+x$ or $-x$. If that goes in the right direction I need somebody to tell me how to proceed.
Assume WLOG that $a_{2n+1}\gt0$. We have $p(x)=a_{2n+1}x^{2n+1}+\dots+a_0$. But, as you take the limit as $x$ approaches $\pm\infty$, the leading term dominates. That is, $\lim_{x\to\pm\infty}p(x)=lim_{x\to\pm\infty}x^{2n+1}(a_{2n+1}+\frac{a_{2n}}{x}+\frac{a_{2n-1}}{x^2} +\dots+\frac{a_0}{x^{2n+1}})=\lim_{n\to\pm\infty}a_{2n+1}x^{2n+1}=\pm\infty$.