Consider the following subset of $P_3(\mathbb{R})$ (real polynomial functions of degree at most 3).
$$ Z = \{f_1, f_2, f_3, f_4, f_5\} $$ where $f_1(x) = 1-2x+2x^2-x3$, $f_2(x) = 1-x+x^2+x^3$, $f_3(x) = 1+3x^3$, $f_4(x) = 1+x+x^2+x^3$ and $f_5(x) = 3-2x+4x^2+x^3$.
Prune $Z$ to produce a linearly independent subset $Y$ with $span(Z) = span(Y)$. What is the dimension of $span(Z)$? Is $p_2$ an element of $span(Z)$? (Recall that $p_2(x) = x^2$.) Extend $Y$ to give a basis for $P_3(\mathbb{R})$.
Write the coefficients of the elements of $Z$ as linear combination of the standard bases $\{1,x,x^2,x^3\}$ as rows in a matrix, then apply the elementary operations on the rows to get the matrix in row echelon form. Then you can read the rank of the matrix, which is equal to the number of independent elements in $Z$. These elements correspond to the rows which are not zero in your final matrix. To see if $p_2 \in \textrm{Span }(Z)$, solve the equation $$p_2= \alpha f_1+\beta f_2+\gamma f_3+ \eta f_4+\zeta f_5$$ for $\alpha, \beta,\gamma,\eta$ and $\zeta$. If it has solution then Yes, If not then No. You can do the same but using the elements of $Y$, because it has less elements than $Z$.
To extend $Y$ to a base of $P_3(\mathbb{R})$ use the final matrix you got from elementary operations, delete the zero rows and add rows so that you make the rank $4$. then add the correspondent elements of the added rows to $Y$.