In Example SP4 we proved that \begin{equation*} W=\setparts{p(x)}{p\in P_4,\ p(2)=0} \end{equation*} is a subspace of $P_4$, the vector space of polynomials of degree at most 4. Since $W$ is a vector space itself, let us construct a span within $W$. First let \begin{equation*} S=\set{x^4-4x^3+5x^2-x-2,\,2x^4-3x^3-6x^2+6x+4} \end{equation*} and verify that $S$ is a subset of $W$ by checking that each of these two polynomials has $x=2$ as a root. Now, if we define $U=\spn{S}$, then Theorem SSS tells us that $U$ is a subspace of $W$. So quite quickly we have built a chain of subspaces, $U$ inside $W$, and $W$ inside $P_4$.
Rather than dwell on how quickly we can build subspaces, let us try to gain a better understanding of just how the span construction creates subspaces, in the context of this example. We can quickly build representative elements of $U$, \begin{equation*} 3(x^4-4x^3+5x^2-x-2)+5(2x^4-3x^3-6x^2+6x+4)=13x^4-27x^3-15x^2+27x+14 \end{equation*} and \begin{equation*} (-2)(x^4-4x^3+5x^2-x-2)+8(2x^4-3x^3-6x^2+6x+4)=14x^4-16x^3-58x^2+50x+36 \end{equation*} and each of these polynomials must be in $W$ since it is closed under addition and scalar multiplication. But you might check for yourself that both of these polynomials have $x=2$ as a root.
I can tell you that $\vect{y}=3x^4-7x^3-x^2+7x-2$ is not in $U$, but would you believe me? A first check shows that $\vect{y}$ does have $x=2$ as a root, but that only shows that $\vect{y}\in W$. What does $\vect{y}$ have to do to gain membership in $U=\spn{S}$? It must be a linear combination of the vectors in $S$, $x^4-4x^3+5x^2-x-2$ and $2x^4-3x^3-6x^2+6x+4$. So let us suppose that $\vect{y}$ is such a linear combination, \begin{align*} \vect{y} &=3x^4-7x^3-x^2+7x-2\\ &=\alpha_1(x^4-4x^3+5x^2-x-2)+\alpha_2(2x^4-3x^3-6x^2+6x+4)\\ &= (\alpha_1+2\alpha_2)x^4+ (-4\alpha_1-3\alpha_2)x^3+ (5\alpha_1-6\alpha_2)x^2\\ &\quad\quad+ (-\alpha_1+6\alpha_2)x+ (-2\alpha_1+4\alpha_2) \end{align*}
Notice that operations above are done in accordance with the definition of the vector space of polynomials (Example VSP). Now, if we equate coefficients, which is the definition of equality for polynomials, then we obtain the system of five linear equations in two variables \begin{align*} \alpha_1+2\alpha_2&=3\\ -4\alpha_1-3\alpha_2&=-7\\ 5\alpha_1-6\alpha_2&=-1\\ -\alpha_1+6\alpha_2&=7\\ -2\alpha_1+4\alpha_2&=-2 \end{align*}
Build an augmented matrix from the system and row-reduce, \begin{equation*} \begin{bmatrix} 1 & 2 & 3\\ -4 & -3 & -7\\ 5 & -6 & -1\\ -1 & 6 & 7\\ -2 & 4 & -2 \end{bmatrix} \rref \begin{bmatrix} \leading{1} & 0 & 0\\ 0 & \leading{1} & 0\\ 0 & 0 & \leading{1}\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix} \end{equation*}
Since the final column of the row-reduced augmented matrix is a pivot column, Theorem RCLS tells us the system of equations is inconsistent. Therefore, there are no scalars, $\alpha_1$ and $\alpha_2$, to establish $\vect{y}$ as a linear combination of the elements in $U$. So $\vect{y}\not\in U$.