Is the set of all polynomials of degree $3$ not a vector space? I am struggling with an example my textbook gave of a set that is not a vector space. It is stated that $V$, the set of all polynomials of degree exactly $3$ is not a vector space. The reason the textbook gives is that this set does not contain a zero vector. However is $f(x) = x^3$ not a polynomial in the set, thus leading to $ f(0) = 0 $ being a zero vector? I understand that this set would still not be a vector space due to a condition like this: $ [1 + 4x^2 + x^3] + [2-x+x^2-x^3] = 3-x+5x^2 $, with the sum not being in the set $V$. However I am still confused about the zero vector example my textbook gives. 
 A: To offer a different perspective and to be a bit nit-picky: 
The zero vector would be the zero polynomial, which is just $0$. The reason why I am pointing this out is because you don't specify the field $F$ over which you are working. In general, polynomials are not the same as polynomial functions, meaning that you "cannot plug in values into polynomials and say that the polynomial $f$ is the same as the collection $(f(x))_{x \in F}$."
For an example, assume that $F$ is the finite field with two elements. Then $f = x^2 + x$ is a polynomial different from $0$. However, if you plug in both elements $0$ and $1$, you get $f(0) = f(1) = 0$. 
If you are working over an infinite field, e.g. the real numbers, then what you do is okay and not at all problematic. In this situation, we can identify polynomials with polynomial functions. (One can show that one can identify polynomials and polynomial functions if and only if the field one is working with is infinite.)
A: The 'zero' in this set would have to be the zero polynomial. Notice $f(x)=x^3$ is only $0$ at $x=0$. So the zero here is the zero function $f(x)= 0$. You want the set of odd degree polynomials to be $\{a_3 x^3 + a_2 x^2 + a_1x+a_0\}$, where not both $a_3$ and $a_1$ are zero so that the polynomial is odd in the traditional sense. 
