Answer
See answer below
Work Step by Step
Let $S$ be the set of polynomial of degree 5 or less whose co-efficients are even integers.
We have $p(x)=4x+6x^2 \in S$. But $\pi p(x)=4\pi x + 6 \pi x^2$ is not in $S$ since its coefficient are not even integers.
Hence, $S$ is not closed under scalar multiplication by vectors, so $S$ is not a vector space over $R$.
