Answer
See below
Work Step by Step
Let $u,v \in S$
$u=a_0+a_1x+a_2x^2:a_0+a_1+a_2=0\\
v=b_0+b_1x+b_2x^2:b_0+b_1+b_2=0$
$\Rightarrow u+v=a_0+a_1x+a_2x^2+b_0+b_1x+b_2x^2\\
=a_0+b_0+(a_1+b_1)x+(a_2+b_2)x^2$
Since $a_0+b_0+a_1+b_1+a_2+b_2=0 \rightarrow u+v \in S$
$\Rightarrow u+v$ is 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 $\lambda \in R$ be a scalar.
$\Rightarrow \lambda.u=\lambda(a_0+a_1x+a_2x^2)=\lambda a_0+\lambda a_1x+\lambda a_2x^2$
Since $\lambda a_0+\lambda a_1x+\lambda a_2x^2=\lambda (a_0+a_1+a_2)=\lambda.0=0$
Therefore $S$ is closed under addition.
