Whether polynomials $(t-1)(t-2),(t-2)(t-3),(t-3)(t-4),(t-4)(t-6)$ are linearly independent. Question is to check if :
$(t-1)(t-2),(t-2)(t-3),(t-3)(t-4),(t-4)(t-6)\in \mathbb{R}[t]$ are linearly independent. 
Instead of writing linear combination and considering coefficient equations, I would like to say in the following way :
set of all polynomials of degree $\leq 2$ is a vector space with basis $1,t,t^2$ over $\mathbb{R}[t]$
 and there are 4 polnomials in the collection$(t-1)(t-2),(t-2)(t-3),(t-3)(t-4),(t-4)(t-6)\in \mathbb{R}[t]$
any collection of $n+1$ elements in a vector space of dimension $n$ is linearly dependent.
Thus, the collection $\{(t-1)(t-2),(t-2)(t-3),(t-3)(t-4),(t-4)(t-6)\}$ is linearly dependent in $\mathbb{R}[t]$
I would be thankful if some one can say if this justification is correct
i would be thankful if someone wants to say something more about this kind of checking.. 

If you assume that the last one is a linear combination of the first three, that is, that we can write
$$
 (t-4)(t-6) = a(t-1)(t-2) + b(t-2)(t-3) + c(t-3)(t-4)
$$
and expand the four polynomials, then you get the three equations
$$
\begin{cases}a + b + c = 0 & \text {from } t^2 \\-3a -5b-7c = -10 & \text{from } t \\ 2a + 6b + 12c = 24 & \text{from } 1\end{cases}
$$
Solving these equations then reveals that $a = \frac{3}{2}$, $b = -\frac{9}{2}$ and $c = 4$, so we get
$$
 (t-4)(t-6) = \frac{3}{2}(t-1)(t-2) -\frac{9}{2}(t-2)(t-3) + 4(t-3)(t-4)
$$
Of course, there were no guarantee that we could write the last polynomial as a linear combination of the others. It could've been the case that $(t-1)(t-2)$ could be written as a linear combination of $(t-2)(t-3)$ and $(t-3)(t-4)$, and that $(t-4)(t-6)$ was linearly independent of the rest of them. If that were the case, then the set of equations above would've had no solution, and we would've had to swap which function we're trying to write as a linear combination of the others.
