Checking whether a polynomial of degree at most 2 encompasses all possible polynomials of degree at most 2? So I found that the span of these three basis polynomials yielded $(a+b+c) + (a-b)x + (a+b)x^2$. The question is whether this spans the space of all polynomials of degree at most 2, and I think the answer is yes, but I'm not sure since I don't know whether there's a polynomial that fails the constraints of $(a+b+c), (a-b), (a+b)$. How do I check this?
Like, I'm pretty sure this polynomial - $(a+b) + (a-b)x + (a+b)x^2$ doesn't span, right? I can tell since the 1st and 3rd coefficients must be the same. Is there a system to check this?
I let $d=(a+b+c)$, so $a-b=d-2b-c$ and $a+b = d -c$ and by eye balling it, I seem good: I can choose $d$ for the constant term, then adjust by $c$ for the $x^2$ term, and then adjust $b$ for the $x$ term. Is this how to do it?
 A: You can translate your question to more standard vector spaces and matrices, and in multiple ways!
First way:
The set $\{(a+b+c) + (a-b)x + (a+b)x^2|a,b,c\in\mathbb R\}$ is the set of all polynomials if and only if, for every $\alpha, \beta, \gamma$, the polynomial $$\alpha + \beta x + \gamma x^2$$ is in the set.
This means that what you are asking boils down to the question

Does the system of equations $$\begin{align}a+b+c=\alpha\\ a-b=\beta\\ a+b=\gamma\end{align}$$ always have a solution?

which, if we define $\mathbf x = \begin{bmatrix}a\\b\\c\end{bmatrix}$, can be translated into the question

Does the equation $A\mathbf x=\mathbf b$ always have a solution if $$A=\begin{bmatrix}1 & 1 & 1\\ 1 & -1 & 0\\ 1 & 1 & 0\end{bmatrix}$$

Second way:
We can view the space of all polynomials as a vector space.
Because
$$(a+b+c) + (a-b)x + (a+b)x^2 = a\cdot(1+x+x^2) + b\cdot(1-x + x^2) + c\cdot 1,$$
we can write the set $$\{(a+b+c) + (a-b)x + (a+b)x^2|a,b,c\in\mathbb R\}$$ as the span of the set $\{1+x+x^2, 1-x+x^2, 1\}$.
Knowing the dimension of the space of polynomials of degree at most $2$ is three, we know that the span of a three-element set will be the whole space if and only if the three elements are linearly independent. Writing the three vectors in the basis $1,x,x^2$, the vectors are $$\begin{bmatrix}1\\1\\1\end{bmatrix},\begin{bmatrix}1\\-1\\1\end{bmatrix},\begin{bmatrix}1\\0\\0\end{bmatrix},$$ and the three vectors are independent if the matrix we get by stacking them up is of full rank, i.e. if the matrix
$$\begin{bmatrix}1&1&1\\1&-1&0\\1&1&0\end{bmatrix}$$ is of full rank.
Note that this is the same matrix as before, and remember, for square matrices, the questions "is the matrix $A$ of full rank?" and "Does the equation $Ax=b$ always have a solution?" are equivalent, so this is the same condition as before!
Third way:
Let $\mathbb P_2[x]$ be the space of polynomials of degree at least $2$. Let $\mathcal A$ be the mapping defined as $$\mathcal A:\mathbb R^3\to\mathbb P_2[x]\\
(a,b,c)\mapsto (a+b+c)+(a-b)x+(a+b)x^2$$
It is easy to see that $\mathcal A$ is a linear map, and that the original set you are observing is precisely the image of $\mathcal A$, that is
$$\mathcal A(\mathbb R^3) = \{(a+b+c) + (a-b)x + (a+b)x^2|a,b,c\in\mathbb R\}.$$ Now, observe the matrix $A$ that is associated with the mapping $A$ in the standard bases of both vector spaces. Note that $$\begin{align}
\mathcal A(1,0,0)&=1+x+x^2\\ 
\mathcal A(0,1,0)&=1-x+x^2\\ 
\mathcal A(0,0,1)&=1
\end{align}$$
which means, by definition, that
$$A=\begin{bmatrix}1&1&1\\1&-1&0\\1&1&0\end{bmatrix}.$$
Now, using some linear algebra, you may remember that a mapping between two equally dimensional spaces is surjective if and only if it is bijective, which happens if and only if the matrix representing it is of full rank, i.e. if the matrix $A$ (which we saw twice before!) is of full rank.
