Find sum of Subspaces combination of constraints Let $ S = \langle \{ x^2-1,x^2+1\} \rangle $
and $ T = \{p\in P_2(\mathbb{R}), p(x)= ax^2+bx+c : a+b=0,c=0\} $ 
$\color{black}{a) \ Find \ S+T}$
From the definition of Sum in subspaces, we take e.g $s_1,t_1$:
S+T = $\{s+t \mid s_1 \in S \wedge t_1 \in T   \}$
And example from the book ''Linear Algebra done right'', in $ \mathbb {R^3}$ state the following :
$U=\{(x,0,0) \in {F^3}:x \in F\}$ and $ W= \{0,y,0) \in F^3:y\in F\} $ 
And then $ U+W = \{x,y,0):x,y \in F \}$ 
And what i think item a) is :
$S+T=\{ax^2+bx+c \mid b=0,a=1,c=1,c=-1\}$
What have i done better?
$\color{black}{b)\ Write \ p(x)=2x^2-5x+6\ as \ a \ sum \ of \ polynomials \ s(x)  \in S} \ and \ a \ polynomial \ t(x) \in T$
Let's say : 
Let's take $s(x) \in S \Rightarrow s(x) = \{x^2-1,x^2+1\}$ 
and $t(x)=ax^2+bx+c \mid a=-b,c=0$
\begin{align}
& = -bx^2-bx+0=0\\
& = -bx^2-bx=0 \\
& = bx(-x-1)=0 \\
& = then , \ x=0\ or\  x=-1 \\
\end{align}
It's supposed to be looking for scalars in $\mathbb{R}$ ?
$\color{black}{c) \ Is \ S \oplus T =P_2(\mathbb{R})?}$ (Direct sum,need to be null, or the solutions of the system need to be infinity?)
I would appreciate some help, thanks.
 A: I think the notation for a vector used in the book is $(a,b,c)$ instead of $\langle a, b, c \rangle$. According to the problem $S$ has two elements: $\{(1, 0, 1), (-1, 0, 1) \}$. By the way, I am using the standard basis $\{1,x,x^2\}$. An element of $T$ can be simplified to $p(x) = ax(x-1)$, and with the notation $(0,-a, a)$.
An element of the sum $S + T$ is $(1, 0, 1) + (-1, 0, 1) + (0, -a, a)$, simplifying $(0, -a, 2 +a)$. So $S + T = \{ p \in P_2(\mathbb R) \mid p(x) = (2+a)x^2 - ax\}$.
To find $p(x) = 2x^2 - 5x + 6$ as sum of one polynomial from $S$ and one polynomial from $T$, write $p(x)$ in terms of the standard basis: $(6, -5, 2)$, then:
$$(6, -5, 2) = \alpha(1, 0, 1) + \beta(-1, 0, 1) + \gamma(0, -a, a)$$
with 
$$\begin{align}
6 &= \alpha - \beta, \\
-5 &= -a\gamma, \\
2 &= \alpha + \beta + a\gamma. 
\end{align}$$
The solution is $\alpha = 3/2$, $\beta=-9/2$, $a\gamma = 5$. I might express the answer differently (check the answer!):
$$6 -5x + 2 x^2 = 3/2\,(1+x^2) -9/2\, (-1 + x^2) + 5(-x + x^2).$$
The last question: Is $S \oplus T = P_2 (\mathbb{R})$? You have to prove (or disprove) that $S \cap T = 0$. Well, you have two elements of $S$ and a way of finding elements of $T$, can you find $t \in T$ such as $t = (1, 0, 1)$ or $t= (-1, 0, 1)$? Can you continue from this?
