<s>[INST] Find $a, b$ so polynomial is divisible 
Find values $a,b\in\Bbb{R}$ so the polynomial $$P(x)=6x^4-7x^3+ax^2+3x+2$$
  is divisible by the polynomial
  $$Q(x)=x^2-x+b$$

So what I know and how I do these problems most of the time, is since:
$$P(x)=Q(x)D(x)$$
I would know that by plugging the roots of $Q(x)$ in $P(x)$ should give me enough equations for me to solve this.
So I tried finding the roots of $Q(x)$:
$$x^2-x+b=0\\x_{1/2}={1\pm\sqrt{1-4b}\over2}$$
Okay so this $b$ value is giving me a headache here. The only thing I gathered from this is that (probably) $b\le0$. I tried now plugging this in $P(x)$, as I know that
$P(x_1)=0$ and $P(x_2)=0$
The exponents on everything made this a real pain and I'm pretty certain that it shouldn't be done this way. I'm stuck.
 [/INST] Simply dividing works:
$$
\require{enclose}
\begin{array}{l}
\phantom{x^2-x+b\,\,\,}6x^2-x+(a-6b-1)\\[-4pt]
x^2-x+b\enclose{longdiv}{6x^4-7x^3+\phantom{6}ax^2+3x+2}\\[-4pt]
\phantom{x^2-x+b\,\,\,}\underline{6x^4-6x^3+6bx^2}\phantom{.00\,00\,00}\\[-4pt]
\phantom{x^2-x+b\,\,\,6x^4}\quad{-x^3+(a-6b)x^2}\\[-4pt]
\phantom{x^2-x+b\,\,\,6x^4}\quad\underline{-x^3+x^2-bx}\\[-4pt]
\phantom{x^2-x+b\,\,\,6x^4}\quad(a-6b-1)x^2+(b+3)x\\[-4pt]
\phantom{x^2-x+b\,\,\,6x^4}\quad\underline{(a-6b-1)x^2-(a-6b-1)x+b(a-6b-1)}\\[-4pt]
\phantom{x^2-x+b\,\,\,6x^4(a-6b-1)x^2\quad\quad}(a-5b+2)x+2+b+6b^2-ab
\end{array}
$$
So we need $a=5b-2$ and $6b^2-ab+b+2=b^2+3b+2=0$.
Thus, $(a,b)\in\{(-7,-1),(-12,-2)\}$.
</s>