Prove that $\{x^2+x , x^2-1, x+1\}$ generates vector space of $2^{nd}$ degree polynomials I am quite new to linear algebra and am having some trouble with the abstractness of some it's parts. For example, this task seems quite simple and as if there's no need to do any proving, but to actually prove it using linear algebra axioma can seem tricky. 
If the set  $\{x^2+x , x^2-1, x+1\}$  generates vector space of $2^{nd}$ degree polynomials, that would have to mean that the linear span of this set is this vector space i.e. 
$L(\{x^2+x , x^2-1, x+1\}) = P_2[\mathbb{R}]$
This means that for some vector $p$ from $P_2[\mathbb{R}]$, :
$$p=\alpha(x^2+x)+\beta (x^2-1) + \gamma (x+1)$$
$$p=x^2(\alpha +\beta ) + x(\alpha + \gamma)+(\gamma-\beta) $$
Now, the real question here is: In what relation do these coefficient with $x^2$, $x$ and $x^0$  have to be in order for this set to generate this vector space. I believe none of them can be zero in order for this to be true. This is a required property, but is it enough?
 A: i want to see if $x$ can be generated by $\{x^2 + x, x^2 - 1,x+1\}.$ suppose  $$a(x^2 + x)+b(x^2 - 1) + c(x+1) = x$$  is an identity. then we have $$a+b = 0,\\a+c =1,\\-b+c= 0 $$ this system of equations gives $$a = -b, c = b, a+c = 0 $$ therefore is inconsistent. that is $\{x^2 + x, x^2 - 1,x+1\}$  cannot generate $x$ and therefore cannot generate second degree polynomials.  
A: The space of polynomials of $\deg 2$ has dimension 3. So, the polynomials in question will generate the space if they are linearly independent.
A: Write the matrix of vectors in the basis $\{1,x,x^2\}$  and use row reduction:
$$\begin{bmatrix}0&-1&1\\1&0&1\\1&1&0\end{bmatrix}\rightsquigarrow\begin{bmatrix}1&0&1\\0&-1&1\\1&1&0\end{bmatrix}\rightsquigarrow\begin{bmatrix}1&0&1\\0&-1&1\\0&1&-1\end{bmatrix}\rightsquigarrow\begin{bmatrix}1&0&1\\0&-1&1\\0&0&0\end{bmatrix}$$
The matrix has rank $2$, just as the polynomials. Hence they are not a basis of polynomials of degree $\le2$.
A: That is wrong becouse  those vectors are linearly depended
$$
1\cdot (x^2+x)+(-1) \cdot (x^2-1)+(-1) \cdot (x+1)=0.
$$
