$f=2X^4-3X^2+aX+b,\ g=X^2-2X+3, \ f,g \in \mathbb{Q}[X]$
I have tried to divide $f$ by $g$ but I get $ (a+10)X +b +3$ as the remainder which looks like a bad result. I have, also, tried to factor $g$, but the only way to factor it is over the complex numbers, and since I am working in $\mathbb{Q}[X]$ I can't do that.
If you divide $f$ by $g$, you get a remainder of $r(x)= (a-14)x+b+3$. That means you can write
$$2x^4-3x^2+ax+b = (x^2-2x+3)(2x^2+4x-1) + r(x)$$
For $f$ to divide $g$, we need $r(x)=0$ for appropriately chosen $a$ and $b$, ie. take $a=14$ and $b=-3$.