Irreducibility in the polynomial ring over the rationals $\mathbb Q[t]$ versus $\mathbb Z[t]$ Context: In this exact moment of this online lecture the following statement is made (consider misunderstandings):

Let $R$ be a ring. An element $a \in R$ is called irreducible if it cannot be written as a product of two elements in $R$ unless one of those elements is a unit.
That is, $a$ is irreducible if $a=bc\implies \left(b\in R^\times \text{ or } c\in R^\times \right).$

Example in the polynomial ring over the rationals:
$p=18 - 12t=(6)(3-2t)$ is irreducible despite factoring out $6.$
Here is the quote:

$(6)$ doesn't count, because it is a unit in $\mathbb Q,$ although it would count over the integers, $\mathbb Z[t].$

My take is that the understanding the speaker is making reference to is the concept of the polynomial ring over the rationals as a vector space with basis $\color{red}1, x, x^2,\dots.$ And I don't know how this translates to the subring of polynomials over the integers, $\mathbb Z[t],$ except the lead in Wikipedia regarding the basis of such polynomials being
$$P_k(t)=t(t-1)\dots(t-k+1)/k!$$
for $k=0,1,2,\dots$ An unfamiliar expression thus far...
 A: It's always the missing steps...
Here is the answer (Wikipedia so often to the rescue): 

In mathematics, an invertible element or a unit in a ring $R$ is any element $u$ that has an inverse element in the multiplicative monoid of $R$, i.e. an element $v$ such that
$uv = vu = 1R,$ where $1R$ is the multiplicative identity.

The multiplicative identity in $\mathbb Q[t]$ is $1;$ hence $6$ is a unit since $1/6 \in \mathbb Q[t].$ Yet, $1/6 \notin \mathbb Z[t].$
A: As others have said, vector spaces are not required here.  
The issue is how to generalise the concept of irreducible or prime to more general rings.  Often, but not always, they are the same.  I will ignore the distinction here since I am just trying to explain the motivation of the definition of unit.  One interesting question in a ring is whether prime factorisation is unique; sometimes it is and sometimes it is not.  
Is prime factorisation unique in $\mathbb{Z}$?  We normally say it is but how about $6 = 2 \times 3 = (-2) \times (-3)$.  Does that violate unique factorisation?  No, because we usually ignore the negatives integers.  Also, $6 = 2 \times 3 = 1 \times 2 \times 3$.  Does that violate unique factorisation?  No, because it is no longer usual to consider $1$ prime.  It is not composite either, so what is it?  It's another category: "unit" - an element with an inverse in the ring.
Now let's look at a more interesting example: the Gaussian Integers, $\mathbb{Z}[i]$.  These are the complex numbers with integer real and imaginary components.  At first, it seems that prime factorisation is not unique since $10 = 2 \times 5 = (3 + i) \times (3 - i)$.  However, the catch is that $2$ is not a prime in this ring since $2 = (1 + i) \times (1 - i)$.  $(1 + i)$ and $(1 - i)$ are prime.  Is the factorisation unique?  Well, $2 = (-1 + i) \times (-1 - i)$ so it seems not.  However, these are just the same factors multiplied by $i$ and $-i$ and these are units, they have inverses in the ring.  These units allow you to create multiple factorisations.  So, it is usual to ignore them; a prime factorisation is considered unique if any other can be obtained from it by rearranging the terms or multiplying them by units.  
Back to the normal integers, the problem can be easily hidden by picking the positive integers as the canonical factors.  This is not so easy in the Gaussian integers or arbitrary rings.  
Now, to your examples.  In $\mathbb{Q}[t]$, all degree zero polynomials are units, they have inverses, and hence are not interesting in factorisations.  But in $\mathbb{Z}[t]$, only the two degree zero polynomials $1$ and $-1$ have inverses, the others might or might not be prime or irreducible.  
Finally, consider a field such as $\mathbb{Q}$ or $\mathbb{R}$.  All elements, except $0$, have an inverse and hence are units.  There are no primes and hence the question of unique factorisation is uninteresting.  
