A question about polynomials as vectors 
The space $P_n(R)$ is very similar to the vector space $R^{n+1}$; indeed one can match
  one to the other by the pairing
$a_nx^n + a_{n−1}x^{n−1} + \ldots + a_1x + a_0$ is equivalent to $(a_n, a_{n−1}, \ldots, a_1, a_0)$,
thus for instance in $P_3(R)$, the polynomial $x^3 + 2x^2 + 4$ would be associated
  with the 4-tuple $(1, 2, 0, 4)$.

I am not sure I understand the equivalence. The left side of it looks like it could be summed up to a scalar - single number, while the right side is a vector in $R^n$. 
The same goes for the cubic polynomial which could be summed up to be a scalar for some $x$ whereas the 4-tuple is a vector in $R^4$.
Please, explain all this.
Thanks. 
 A: In an arbitrary $n$-dimensional vector space with basis $\{\mathbf{f}_1,\ldots,\mathbf{f}_n\}$, it is usual to denote by the $n$-tuple $(x_1,\ldots,x_n)$ the vector 
$$\sum_{i=1}^n x_i\mathbf{f}_i$$
If one fixes the following basis $\{1,x,\ldots,x^n\}$ of $P_n(\mathbb R)$, then the polynomial $x^3+2x^2+4$ can be written as
$$\mathbf{1}\cdot x^3 + \mathbf 2\cdot x^2 + \mathbf 0\cdot x +\mathbf 4 \cdot 1$$
Which (by the above) is represented by the tuple $(1,2,0,4)$, which we can identify with an element of $\mathbb R^{3+1} = \mathbb R^4$.
The explicit isomorphism is given by mapping the basis vector $x^i \in P_n(\mathbb R)$ to the $(n-i+1)$-th standard basis vector $e_{n-i+1}\in \mathbb R^{n+1}$.
The key thing here is that the powers of $x$ are not scalars, they are indeterminates in a polynomial ring. You should check that adding polynomials and multiplying them my scalars under this identification with an $(n+1)$-tuple works the same as in $\mathbb R^{n+1}$ if you're not yet convinced, but explicitly constructing (and checking) the map I defined above is an isomorphism of vector spaces should convince you.
A: The space of polynomials, the original algebra, can be defined as the space of finite infinite sequences, where "finite" refers to the property that only a finite number of terms is different from zero.
The evaluation of a polynomial at some given point gives a concrete number as result, this gives a polynomial function if a domain for these evaluations is fixed.
But in the formula above, $x$ is an indeterminate, a formal symbol that denotes in its power the place in the sequence for the coefficient before it.
A: As you already stated, this is an equivalence, no equation. So you actually can't set both sides as equal.
If you define a variable $x$ in $\mathbb{R}$, you can identify it for example as the tuple $(x,0)$ in $\mathbb{R}^2$. You have an equivalence between both elements, but they are not equal, as one lies in $\mathbb{R}$ and the other in $\mathbb{R}^2$. You can define a function 
$$
f: \mathbb{R} -> \mathbb{R}^2  \\
f(x) = (x,0)
$$
that identifies one element with the other. Can you define a function that identifies the polynomial elements with the vectors in your equivalence? Or even an inverse function?
