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.
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.
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 

$$
f: \mathbb{R} -&gt; \mathbb{R}^2  \\
f(x) = (x,0)
$$

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?