Answer
See below
Work Step by Step
Given $V=P_3(R)\\
S=\{2x-x^3,1+x+x^2,3,x\}$
Let $a,b,c$ be scalars
Obtain $a(2x-x^3)+b(1+x+x^2)+3c+dx=0\\
2ax-ax^3+b+bx+bx^2+3c+dx=0\\
(3c+b)+(2a+b+d)x+bx^2-ax^3=0$
We have the system $3c+b=0\\
3a+b+d=0\\
b=0\\
-a=0$
Substitute: $3c+0=0\rightarrow c=0\\
2(0)+0+d=0\rightarrow d=0$
We can see that $a=b=c=d=0$, then the linear combination of $S$ is trivial. Hence, $S$ is a linearly independent set in $P_3(R)$
We have $dim[P_3]=4$ and $S$ is a linearly independent set in $P_3(R)$, hence $S$ spans $P_3(R)$
