There is a linear map on $\rm \mathbb{P}_n(\mathbb{R})$ given by $\rm L_a :f(x)\to f(x+a)$ for any $\rm a\in\mathbb{R}$. It is invertible because we can see $\rm L_a L_{-a}=L_{-a}L_a=Id$ as linear maps. Hence $\rm L_a$ preserves dimensions of subspaces. We may then write $\rm L_{-a} W=V=\{ p(x)\in \mathbb{P}_n(\mathbb{R}):p(0)=0\}$. It is easy to see that $\rm B=\{x,x^2,\dots,x^n\}$ is a basis for this space by showing that $\rm p(0)=0\,$ is equivalent to $\rm p_0=0$ (the constant term vanishes).
One may use this to show $\rm L_{-a}B=\{x-a,(x-a)^2,\dots,(x-a)^n\}$ is a vector space basis for $\rm W$.
Write your four vectors as column vectors of a $ \ 5 \times 4 \ $ matrix and row reduce it:
$$ \left( \begin{array}{cc} 6 & 1 & 1 &7 \\4&0&4&1\\1&2&-9&0\\-1&3&-16&-1\\2&-4&22&3 \end{array} \right) \ \ \rightarrow \ \ \left( \begin{array}{cc} 0 & 0 & 0 &0 \\0&0&0&0\\0&0&0&1\\1&-3&16&1\\0&1&-5&\frac{1}{2} \end{array} \right) $$
[I used the fourth row here to work against the other rows; that doesn't matter particularly.]
What does the "zeroing-out" of two rows tell us? How can we use what non-zero rows remain to construct a basis for span(S) ? (Notice that these are five-dimensional vectors, so we are already starting out "short a coordinate variable", making it "free".)
EDIT -- Since the discussion has advanced further, we can say something about the basis of span(S). Taking the hint from Omnomnomnom or the above, the subspace spanned by your set of four vectors only has dimension 3. So we need to set up three linearly independent vectors, using the columns of the row-reduced matrix.
We could "reduce" those last two rows a bit more to obtain
$$ \rightarrow \ \ \left( \begin{array}{cc} 0 & 0 & 0 &0 \\0&0&0&0\\0&0&0&1\\1&0&1&0\\0&1&-5&0 \end{array} \right) \ \ . $$
With the matrix fully "reduced", we need to pick out three (five-dimensional) column vectors which are linearly independent. The third column is a linear combination of the first two, so we can toss that one out. A suitable basis for span(S) is then
$$ \left( \begin{array}{cc} 0 \\0\\1\\0\\0 \end{array} \right) \ \ , \ \ \left( \begin{array}{cc} 0 \\0\\0\\1\\0 \end{array} \right) \ \ , \ \ \left( \begin{array}{cc} 0 \\0\\0\\0\\1 \end{array} \right) \ \ . $$
Best Answer
Hint: What is the degree of the polynomial once you factor the root $(x-a)$ from $f$? If $a$ is the only root, then $g(x)(x-a) = f(x)$ and $g(x)$ would be a basis.
The reason that the set $1, x, x^2, ..., x^n$ is a basis for $P_n(\mathbb{R})$ is because by definition a polynomial $p \in P_n(\mathbb{R})$ has the form: $$p(x)= a_n x^n+a_{n-1}x^{n-1}+...a_0$$ thus a linear combination of the selected basis hits all vectors in the space. We need not choose this set as a basis, however. Any linearly independent set of vectors which has a length corresponding to the dimension of the space will also be a basis.