Why is absolute value function not a polynomial? 
Why is absolute value  function   not a polynomial?

I need  a clear  answer to this question please,?
Why couldn't we consider absolute value function as a polynomial? 

Sorry for the late answer, but here is one without differentiability or the definition:
Lemma 1: $f$ and $g$ and polynomials, then $\deg (f * g) = \deg f + \deg g$
Proof:
From the definition of the degree of a polynomial:
$$\deg f=\deg f'$$
$$\deg g=\deg g'$$
where $f'=a_nx^n$, $g'=b_mx^m$ ($n=\deg f$, $m=\deg g$).
Then $\deg(f' * g')=\deg(f * g)$ because $f'$ and $g'$ are the summands with the largest power in their respective polynomials. But $f'*g'=a_nb_mx^{m+n}$. Thus $\deg (f' * g')=\deg f'+\deg g'$ and $\deg (f*g)=\deg(f' * g')=\deg f'+\deg g'=\deg f+\deg g$. $\blacksquare$
Note the special case of Lemma 1 for $f=g$, we have $\deg(f^2)=2deg(f)$.
Now, $|x|^2$ is equal to $x^2$. We will prove by contradiction that $|x|$ cannot be a polynomial. The proof is very simple: because $\deg (|x|^2)=\deg(x^2)=2$, $\deg |x|$ would be $1$. This would mean that $|x|$ is a linear function.
But we know that, given two points, there is exactly one line through them (Postulate III). Thus, if we have $|0|=0$ and $|1|=1$, there is exactly one line through these points, and we see $y=x$ matches the criterion and again it is the only one. Thus, $|x|=x$.
Now, recall that $|-1|=1$. But $|-1|=-1$, because $|x|=x$ as proven above. Thus $1=-1$, which is an obvious contradiction. Because $|x|$ being a polynomial leads to contradiction, $|x|$ is definitely not a polynomial. $\blacksquare$
