Finding a monic quadratic polynomial which is a factor of two other polynomials Problem statement:
Find a monic quadratic polynomial, $f(x)$, which divides both $$g(x) = 12x^3 − 30x^2 + 18x − 12$$ and $$h(x) = 6x^4 + 3x^3 + 6x^2 + 3$$
My take on it:
I divided $h(x)$ by $g(x)$ to get the quotient and remainder such that $$6x^4 + 3x^3 + 6x^2 + 3 = (12x^3 − 30x^2 + 18x − 12)(\frac{1}{2}x + \frac{3}{2}) + 3(14x^2 - 11x + 7)$$
It is also the case that any polynomial divisor of both $g(x)$ and $h(x)$ must also divide the remainder polynomial when $h(x)$ is divided by $g(x)$.
So following on from that, our common factor for $g(x)$ and $h(x)$ that we're trying to find, would also have to be a factor of our remainder, $3(14x^2 - 11x + 7)$. Yet the remainder cannot be factorised any further to turn it into a monic quadratic polynomial.
How should I go about solving this problem?
So far, it's only this particular problem that I find myself unable to solve, other ones like it are fine.
Important: Please don't provide the full solution but rather only the way forward as I don't want to be accused of plagiarism. This exact problem is part of an assignment.

EDIT: I actually had an error in my calculation, the resulting equation is now as follows
$$
6x^4 + 3x^3 + 6x^2 + 3 = (12x^3 − 30x^2 + 18x − 12)(\frac{1}{2}x + \frac{3}{2}) + 42(x^2 - \frac{1}{2}x + \frac{1}{2})
$$
 A: we have:
$$
12x^3-30x^2+18x-12=6(2x^3-5x^2+3x-2)
$$
and we see that $x=2$ is a root, so, dividing by $(x-2)$ we find:
$$
12x^3-30x^2+18x-12=6(x-2)(2x^2-x+1)
$$
Now show that $2x^2-x+1$ divides the other polynomial. And you can write it as a monic polynomial  using:
$$
2\left(x^2-\frac{1}{2}x+\frac{1}{2} \right)
$$
A: =-=-=-=-==-
$$  \left(  6 x^{4}  + 3 x^{3}  + 6 x^{2}  + 3 \right)  $$ 
$$  \left(  12 x^{3}  - 30 x^{2}  + 18 x  - 12 \right)  $$ 
$$  \left(  6 x^{4}  + 3 x^{3}  + 6 x^{2}  + 3 \right)  =  \left(  12 x^{3}  - 30 x^{2}  + 18 x  - 12 \right)  \cdot \color{magenta}{  \left(   \frac{  x  + 3 }{ 2 }  \right) } +  \left(  42 x^{2}  - 21 x  + 21 \right)  $$ 
 $$  \left(  12 x^{3}  - 30 x^{2}  + 18 x  - 12 \right)  =  \left(  42 x^{2}  - 21 x  + 21 \right)  \cdot \color{magenta}{  \left(   \frac{ 2 x  - 4 }{ 7 }  \right) } +  \left( 0 \right)  $$ 
 $$ \frac{ 0}{1} $$ 
 $$ \frac{ 1}{0} $$ 
 $$ \color{magenta}{  \left(   \frac{  x  + 3 }{ 2 }  \right) }  \Longrightarrow  \Longrightarrow  \frac{  \left(   \frac{  x  + 3 }{ 2 }  \right) }{ \left( 1  \right) } $$ 
 $$ \color{magenta}{  \left(   \frac{ 2 x  - 4 }{ 7 }  \right) }  \Longrightarrow  \Longrightarrow  \frac{  \left(   \frac{  x^{2}  +  x  + 1 }{ 7 }  \right) }{ \left(   \frac{ 2 x  - 4 }{ 7 }  \right) } $$ 
 $$  \left(   x^{2}  +  x  + 1 \right)  \left( \frac{ 1}{7 } \right)  -  \left(   x  - 2 \right)  \left(   \frac{  x  + 3 }{ 7 }  \right)  =  \left( 1  \right)  $$ 
 $$  \left(  6 x^{4}  + 3 x^{3}  + 6 x^{2}  + 3 \right)  =  \left(   x^{2}  +  x  + 1 \right)  \cdot \color{magenta}{  \left(  6 x^{2}  - 3 x  + 3 \right) } +  \left( 0 \right)  $$ 
 $$  \left(  12 x^{3}  - 30 x^{2}  + 18 x  - 12 \right)  =  \left(   x  - 2 \right)  \cdot \color{magenta}{  \left(  6 x^{2}  - 3 x  + 3 \right) } +  \left(  6 x^{3}  - 15 x^{2}  + 9 x  - 6 \right)  $$ 
 $$  \mbox{GCD} =   \color{magenta}{  \left(  6 x^{2}  - 3 x  + 3 \right) }   $$ 
 $$  \left(  6 x^{4}  + 3 x^{3}  + 6 x^{2}  + 3 \right)  \left( \frac{ 1}{7 } \right)  -  \left(  6 x^{3}  - 15 x^{2}  + 9 x  - 6 \right)  \left(   \frac{  x  + 3 }{ 7 }  \right)  =  \left(  6 x^{2}  - 3 x  + 3 \right)  $$ 
