This is what I have now.
$$
\begin{array}{l}
\mathrm{W}^{\perp}=\{p \in P 2 |\langle p, x+1\rangle=0\} \\
\langle p x+1\rangle=p(-1) x+1(-1)+p(0) x+1(0)+p(1) x+1(1)=p(-1) \\
(-1+1)+p(0)(0+1)+p(1)(1+1)=p(0)+2 p(1)=2 \mathrm{p}(1)
\end{array}
$$
since we are looking for polynomials such that $\mathrm{p}(0)=2 \mathrm{p}(1),$ and with the definition of $\mathrm{P}^1$ all
polynomials $a x^2+b x+c$ such that $c=2(a+b+c),$ so the numbers a,b,c with 2a+2b+c=0. In terms of linear algebra and the null space of $A=[2,2,1]$ which is dimension 2 and generates the vectors
$\begin{bmatrix}1\\0\\-2\end{bmatrix}$ and $\begin{bmatrix}0\\1\\-2\end{bmatrix}$
Which converts back into polynomials to get
$W^{\perp}=\left\{\mathrm{x}^2-2, \mathrm{x}-2\right\}$
Did I solve this question correctly?
