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.. 
 A: That is very good reasoning and much better then actually finding a linear dependence if you do not need one.
Small things: The “over $\mathbb R[t]$” part is funny. I think you want to say something like that the polynomials of degree at most $2$ form a $3$-dimensional subspace of $\mathbb R[t]$. By the way, I think it is good that you give a basis, because that justifies the claim that this is a $3$-dimensional subspace. And while this is implicit, you never actually say that the four polynomial at hand are of degree $2$. 
A: I would say that your reasoning is correct. As you say, you can't have $n+1$ linearly independent vectors in a space of dimension $n$.
A similar "counting" argument that is also useful, sometimes ... in a space of dimension $n$, a set of $n-1$ (or fewer vectors) can not span the space.
A: 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.
A: Let $p_{i}(x)$ (where $1\leq i \leq n$) be $n$ polynomials in $\mathbb{R}[t]$. For checking of linearly dependently or independently these polynomials, in general we can using the following method. Let $d=Max_{1\leq i \leq n} deg(p_{i}(x))$ and $V$ be subspace of all polynomials of degree$\leq d$, then there exits subspace $W$ such that $$\mathbb{R}[t]=V\oplus W,$$
where also can prove that $W\cong \mathbb{R}[t]/V$ and $dim(\mathbb{R}[t])=dim(V)+dim(W)$, in which $dim(V)=d+1$.
Now we can say that the set $\{p_{i}(x) :  1\leq i \leq n \}$ with $n>d+1$ is linearly dependent set, in which $d=Max_{1\leq i \leq n} deg(p_{i}(x))$.
