Express $18X^2-12X+48$ in $\mathbb{Q}[X]$ as the product of a constant polynomial and a primitive polynomial I believe I get the correct answer for the following problem but the solution says it is wrong. I have no idea why.

Express $18X^2-12X+48$ in $\mathbb{Z}[X]$ and in $\mathbb{Q}[X]$ as the product of a constant polynomial and a primitive polynomial.

My work:
In $\mathbb{Z}[X]$ and $\mathbb{Q}[X]$: $6(3X^2-2X+8)$, $6(3X^2-2X+8)$ is primitive because 3,-2,8 are relatively prime in $\mathbb{Z}$ and $\mathbb{Q}$.  
But the solution says in $\mathbb{Q}[X]$ it is just $1(18X^2-12X+48)$. 
I am really confused, if 18,-12,48 are not relatively prime in $\mathbb{Z}$ then they are automatically not relatively prime in $\mathbb{Q}$. Then how can $18X^2-12X+48$ be primitive?
I am pretty sure their solution is correct. Could anyone correct my mistakes?
Thanks!
 A: What prime factors could $18$ and $-12$ have in common? They can't, for example, have $2$ as a common prime factor, because $2$ isn't prime in $\mathbb{Q}$! Instead, it's a unit, like $-1$.
The linear combination definition of relatively prime is even easier to see: there are lots of ways to write $1$ as a linear combination of $18$ and $-12$, such as
$$ \frac{1}{18} \cdot 18 + 0 \cdot (-12) = 1 $$
$$ \frac{1}{9} \cdot 18 + \frac{1}{12} \cdot (-12) = 1$$
A: Ideas like 'relative primality' and 'primitiveness' are, in general, only defined up to multiplication by units.  For example, in $\mathbb{Z}[X]$, we could write $18X^2-12X+48$ as the product of a constant polynomial and a primitive polynomial as 
$$
6(3X^2-2X+8)
$$
But we could also write it as
$$
-6(-3X^2+2X-8)
$$
The thing is, we like to treat these two factorizations as 'basically the same', because the second one is just the first one, multiplied by $-1$.  $-1$ is special, because it has a multiplicative inverse: $-1.-1=1$.  We call elements of a ring that have multiplicative inverses units.  In this case, if we want to ensure uniqueness of the representation, we might specify that the constant polynomial has to be a positive constant.  
In $\mathbb{Q}$, every non-zero number has a multiplicative inverse.  So we can get 'basically the same' factorization by multiplying the constant by $\frac16$ and the primitive polynomial by $6$ to get 
$$
1(18X^2-12X+48)
$$
Why do we say that $18X^2-12X+48$ is primitive?  For the same reason that we say that the polynomial $X^2-X+1$ is primitive in $\mathbb{Z}[X]$: by convention, we treat units like $1$ and $-1$ as empty products (recall that prime factorization is only defined up to multiplication by a unit).  So they have no common factors, since they have no factors at all.  
In $\mathbb{Q}$, none of the numbers have any factors at all, because they're all already units (they have multiplicative inverses).  So they're all automatically relatively prime.  
Now, it's not incorrect to say that 
$$
6(3X^2-2X+8)
$$
is a factorization into a constant polynomial and a primitive polynomial in $\mathbb{Q}[X]$.  However, since we can multiply the constant polynomial by anything we like, divide the primitive by the same number and preserve primitivity of the second polynomial, it's easier to specify that the constant polynomial should be $1$, so we can ensure uniqueness of the representation.  
A: The problem lies in your intuition that coefficients not coprime in $\Bbb Z$ aren't coprime in $\Bbb Q$.
The statement that the coefficients are coprime amounts to the ideal generated by them being the entire ring: $(3,-2,8)=\Bbb Z$. But in $\Bbb Q$, any nonzero thing is going to generate all of $\Bbb Q$, so $(3,-2,8)$ and $(18,-12,48)$ are both all of $\Bbb Q$, so the coefficients are still "coprime."
But in $\Bbb Z$, $(18,-12,48)=6\Bbb Z\neq \Bbb Z$, so that triplet is not a set of coefficients of a primitive polynomial. This is a counterexample to your claim that "not coprime in $\Bbb Z$ implies not coprime in $\Bbb Q$."
I think the upshot of all of this is that your answer of $6(3x^2-2x+8)$ is perfectly fine for both $\Bbb Z$ and $\Bbb Q$. They probably put in the alternative answer to get your mental juices flowing about why $(18^2-12x+48) $ is also right for $\Bbb Q$.
A: Another perspective is simply to notice that $6$ is a unit in $\mathbb Q$ because $\frac 16 \times 6=1$
