Dimension of vector subspace Let $V$be the vectorspace of all polynomials. And $W$ is the subspace spanned by $t^2+t+2, t^2+2t+5, 5t^2+3t+4,2t^2+2t+4$. 
The question asks us to find the dimension of $W$. Here is my try:
Let us call the four polynomials as  $A$,$B$,$C$&$D$ in the order as they appear in the question. Then a cursory look at the four polynomials shows that $A$and $D$ are linearly dependent. So one of them could be removed and we could proceed with remaining $3$. I removed $D$.
 Next when I formed a $3\times3$ matrix having coefficients of $t^2,t$ and constant terms in  $A$,$B$&$C$  as its rows, I found the three polynomials to be linearly dependent. $C$ could be written as a linear combination of $A$ and $B$. So I concluded that $W$ could be spanned by $A$ and $B$.
 But what about the dimension of $W$? How to find that?

The dimension is at most $3$, since $W\subset P_2$.  It is at least $2$, as the first two are not multiples of each other.   The last is twice the first,  so can be thrown out. 
Let's check if the first three are independent, by computing the following determinant:   $\begin{vmatrix}1&1&2\\1&2&5\\5&3&4\end{vmatrix}=1\cdot(8-15)-1\cdot (4-25)+2((3-10)=-7+21-14=0$.
Thus the first three are indeed dependent. 
Thus $\operatorname {dim}W=2$.
