Is constant a polynomial Is constant number a polynomial?

Is $5$ a polynomial?
My present understanding is no:
$0$ is a polynomial with no degree be defined,
But if $5$ is a polynomial, then $5$'s degree is $0$
 A: The algebra of polynomials $k[X]$ over a field $k$ can be built as the set of sentences $k^{(\mathbb{N})}$ (that is the sentences eventually zero) endowed with the following operations:


*

*$(a_n)+(b_n)=(a_n+b_n)$,

*$(a_n) \times (b_n)=(c_n)$ with $\displaystyle c_n= \sum\limits_{i+j=n} a_ib_j$,

*$\lambda \cdot (a_n)=(\lambda \cdot a_n)$.


As a $k$-vector space, we can see that $\{X^0,X^1,X^2,\dots\}$ is a basis of $k[X]$ where $X=(0,1,0,\dots)$ (we suppose that $X^0=(1,0,\dots)$). Therefore, every polynomial can be written as $\displaystyle \sum\limits_{n \geq 0} a_nX^n$ where $(a_n)$ is eventually zero.
In particular, $k$ is canonically embedded into $k[X]$ by $a \mapsto (a,0,\dots)$. 
The degree of a polynomial $P=(a_n)$ is defined as $\deg(P)= \max\{ n \geq 0 \mid a_n \neq 0\}$ if $P \neq 0$; otherwise, $\deg(P)=-\infty$.

Therefore, in $\mathbb{R}[X]$ for example, $5$ is a polynomial of degree zero and $0$ is a polynomial of degree $- \infty$.
Remark: It is convenient to suppose $\deg(0)=-\infty$ so that relations like $\deg(PQ)=\deg(P)+\deg(Q)$ still hold when $P$ or $Q$ is zero.
A: The degree of 0, as far as I know, is set equal to $-1$ (some practical reasons) and constants have degree $0$.
