Is this a correct proof for a span? $V = P_2(\mathbb{R})$ with degree less than 3. Let
$$U = \{f \in V \mid f(2) = 0\} $$
So to prove that$ (t^2-4,t-2) .$ is a basis. Could I do this?
Let $a_1(t^2-4)$ +$ a_2(t-2) $ = $a^2t +bt +c$ and then make a,b and c subjects of $a_1$ and $a_2$ and therefore for any a,b and c you have a combination
 A: It's true that any linear combination of $t^2-4$ and $t-2$ is an element of $U$, but you have to prove also the converse.
So, let $at^2+bt+c\in U$; you want to show that $at^2+bt+c=a_1(t^2-4)+a_2(t-2)$ has a solution, which is possibly what you mean in the last paragraph.
The equations become
\begin{cases}
a_1=a\\
a_2=b\\
-4a_1-2a_2=c
\end{cases}
and the linear system has the matrix
$$
\begin{bmatrix}
1 & 0 & a\\
0 & 1 & b\\
-4 & -2 & c
\end{bmatrix}
$$
A row echelon form for this matrix is
$$
\begin{bmatrix}
1 & 0 & a\\
0 & 1 & b\\
0 & 0 & 4a+2b+c
\end{bmatrix}
$$
Since the polynomial belongs to $U$, we have $4a+2b+c=0$ and so the system has unique solution. Therefore the two vectors form a basis.
A: First show that $t^2 - 4 \in U, t-2 \in U$, which are both clear.
Then show that these are linearly independent, which shouldn't be too hard.
Suppose that $f(t) = at^2 + bt + c$ is in $U$. So 2 is a root of $f$. This means $f(t)$ is divisible by $(t-2)$, so $f(t) = p(t)(t-2)$ where $p(t)$ is linear at most.
Write $p(t) = d(t+2) + e$ (which can be done for any linear $p(t)$. Then $$f(t) = (t-2)(d(t+2) + e) = d(t^2 - 4) + e(t-2)$$
This shows that the two polynomials span $U$. 
