Degree of $P(x)$ satisfying $ \vert P(x) \vert \leq C(1+\vert x \vert)^{5/2}$

Question

Let $P(x)$ be a polynomial in $x$ such that $$ \vert P(x) \vert \leq C(1+\vert x \vert)^{5/2}$$

for all $x \in \mathbb{R}$ and for some constant $C>0.$ Then

1) $P$ is always linear.

2) $P$ is of degree at most 2.

3) $P$ is either a quartic polynomial or cubic polynomial.

4) $P(0)$ is always $C$.

I think option 2 is correct, my idea comes from Liouville's theorem in complex analysis. If option 2 is true then how can I prove? Thanks.

Answer 1

Hint: Consider $$ \lim_{x\to\infty}\frac{P(x)}{(1+x)^{5/2}}\le C $$

Answer 2

Suppose that $\deg P(x)=n>2$. Then you can write $P(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_0$ (with $a_n\neq0$) and$$\lim_{x\to+\infty}\frac{\bigl|P(x)\bigr|}{\bigl(1+|x|)\bigr)^\frac52}=\lim_{x\to+\infty}\frac{\bigl|a_nx^n+a_{n-1}x^{n-1}+\cdots+a_0\bigr|}{\bigl(1+|x|)\bigr)^\frac52}=+\infty$$since $|a_n|\neq0$, thereby reaching a contradiction; this quotient should be bounded by $C$.