Answer
See below
Work Step by Step
Let take two elements from $S$:
$p(x)=x^2+x$ and $q(x)=-x^2$ :
$\Rightarrow p(x)+q(x)=x^2+x+(-x^2)=x$
Hence $p(x)+q(x) \notin S$.
$\Rightarrow S$ is not closed under addition.
Set of Rational numbers is closed under usual addition.
Because our scalars can come from set of real numbers
In particular, let $\lambda=0$ be a scalar.
and $p(x)=x^2+x \in S$
Obtain: $\lambda .p(x)=0.(x^2+3)=0$
Since degree of $\lambda. p(x)$, hence $\lambda.p(x) \notin S$
Therefore $S$ is not closed under scalar multiplication.
