$\mathscr{P}_{2}(\mathbb{R})$ is the vector space of degree 2 polynomials in real coefficients.
What I did so far:
$v \in S=span(1+x,x+2x^2,1-x^2) \iff \exists \ \alpha_{1},\ \alpha_{2},\ \alpha_{3} \in \mathbb{R} : \ v=\alpha_{1}(1+x)+\alpha_{2}(x+2x^2)+\alpha_{3}(1-x^2)$
To show that $\mathscr{P}_{2}$ is in the span of the given set is the same as showing that $\exists \ \beta_{1},\beta_{2},\beta_{3} \in \mathbb{R}$ such that $v=\beta_{1}+\beta_{2}x+\beta_{3}x^2 \in S$, right? I obtained the following linear system from the $v$ vector above:
$(\alpha_{1}+\alpha_{3})1+x(\alpha_{1}+\alpha_{2})+x^2(2\alpha_{2}-\alpha_{3})=\beta_{1}+\beta_{2}x+\beta_{3}x^2$
It can't be solved for the $\alpha$'s, is that sufficient to conclude the set can't generate $\mathscr{P}_{2}$?
Since the determinant of $$\begin{vmatrix}1&0&1\\1&1&0\\0&2&-1\end {vmatrix}=1\neq0.$$
You are incorrect. The vectors are independent.