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
$$
6x^4 + 3x^3 + 6x^2 + 3 = (12x^3 − 30x^2 + 18x − 12)(\frac{1}{2}x + \frac{3}{2}) + 42(x^2 - \frac{1}{2}x + \frac{1}{2})
$$

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:
$$
2\left(x^2-\frac{1}{2}x+\frac{1}{2} \right)
$$
