$\operatorname{Span}\{f_1,f_2,f_3,f_4,f_5\}=\mathbb{R}_4[x]$ I have 
$$f_1=1+x+x^2+x^3$$
$$f_2=x+x^2+x^3+x^4$$
$$f_3=x+2x^2+x^4$$
$$f_4=1+3x+4x^2+2x^3$$
$$f_5=1+x^4$$
If I want to show $\operatorname{Span}\{f_1,f_2,f_3,f_4,f_5\}=\mathbb{R}_4[x]$, I need to show that $\forall g\in \mathbb{R}_4[x]$,
$$g=af_1+bf_2+cf_3+df_4+ef_5, a,b,c,d,e\in \mathbb{R}$$
where $g=\alpha+\beta x+\gamma x^2+\delta x^3+ \varepsilon x^4, \alpha,\beta,\gamma,\delta,\varepsilon\in \mathbb{R}$
So by distribution,
$$(a+d+e)+(a+b+c+3d)x+(a+b+2c+4d)x^2+(a+b+2d)x^3+(b+c+e)x^4$$
But now how do I show that
$$a+d+e=\alpha$$
$$a+b+c+3d=\beta$$
$$a+b+2c+4d=\gamma$$
$$a+b+2d=\delta$$
$$b+c+e=\varepsilon$$
yields all real coefficients for polynomials in $\mathbb{R}$?
 A: Hint:
Your system has a unique solution for any $(\alpha,\cdots,\varepsilon)$ iff the matrix of coefficients is full rank, that is the determinant is nonzero.
Another ad hoc method:
Since $f_1-f_2=1-x^4$ together with $f_4=1+x^4$ span $\{1,x^4\}$. Your problem reduce to $\bar{f}_2,\bar{f}_3,\bar{f}_4$ span $\{\bar{x}, \bar{x}^2,\bar{x}^3\}$, where $\bar{x}$ means modulo $<1,x^4>$.
A: I'm assuming that $\Bbb R_4[x]=\DeclareMathOperator{Span}{Span}\Span\{1,x,x^2,x^3,x^4\}$.
Since $\dim \Bbb R_4[x]=5$, it suffices to determine if $\{f_1,\dotsc,f_5\}$ are linearly independent. To do so, note that
$$
\lambda_1 f_1+\dotsb+\lambda_5 f_5=\mathbf 0
$$
if and only if
\begin{align*}
\lambda_1+\lambda_4+\lambda_5 &= 0 \\
\lambda_1+\lambda_2+\lambda_3+3\lambda_4 &= 0 \\
\lambda_1+\lambda_2+2\lambda_3+4\lambda_4 &= 0 \\
\lambda_1+\lambda_2+2\lambda_4 &= 0\\
\lambda_2+\lambda_3+\lambda_5 &=0
\end{align*}
or equivalently
$$
A\overrightarrow{\lambda}=\mathbf 0
$$
where 
$$
A=
\begin{bmatrix}
1 & 0 & 0 & 1 & 1 \\
1 & 1 & 1 & 3 & 0 \\
1 & 1 & 2 & 4 & 0 \\
1 & 1 & 0 & 2 & 0 \\
0 & 0 & 1 & 0 & 1
\end{bmatrix}
$$
But 
$$
\operatorname{Null}(A)=\Span\left\{\begin{bmatrix}-2\\0\\-1\\1\\1\end{bmatrix}\right\}
$$
This implies
$$
-2f_1-f_3+f_4+f_5=\mathbf 0
$$
so that $\{f_1,\dotsc,f_5\}$ are not linearly independent. Hence $\Span\{f_1,\dotsc, f_5\}\neq \Bbb R_4[x]$. In fact, the above shows that $\Span\{f_1,\dotsc, f_5\}$ is a four-dimensional subspace of $\Bbb R_4[x]$.
A: This is not an answer (well I guess it is) just verification
The Matrix of coefficients then is
$$\begin{bmatrix}a&0&0&d&e\\a&b&c&3d&0\\a&b&2c&4d&0\\a&b&0&2d&0\\0&b&c&0&e\end{bmatrix}$$
Applying Row operations $R_3-R_2\rightarrow R_3$ and $R_2-R_4\rightarrow R_4$, I get
$$\begin{bmatrix}a&0&0&d&e\\a&b&c&3d&0\\0&0&c&d&0\\0&0&c&d&0\\0&b&c&0&e\end{bmatrix}$$
Since $R_3=R_4, Span\{f_i,i=1..5\}\neq\mathbb{R}_4[x]$
