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.
 A: Let $D(x) = 6x^2+cx+d$ (to fix the coefficient of $x^4$). Then multiply out:
$$Q(x)D(x) = (x^2-x+b)(6x^2+cx+d) = 6x^4+(c-6)x^3+(6b+d-c)x^2+(bc-d)x+bd = 6x^4-7x^3+ax^2+3x+2$$
Equate coefficients:
$$c-6=-7 \Longrightarrow c=-1$$
$$6x^4-7x^3+(6b+d+1)x^2-(b+d)x+bd = 6x^4-7x^3+ax^2+3x+2$$
$$bd = 2 \Longrightarrow d=\dfrac{2}{b} \\ b+d=-3 \Longrightarrow b+\dfrac{2}{b}=-3 \Longrightarrow b^2+3b+2=0 \Longrightarrow b\in \{-1,-2\}$$
Case 1: $b=-1$
$$d=\dfrac{2}{b}=\dfrac{2}{-1}=-2 \\ 6b+d+1 = 6(-1)+(-2)+1 = -7 = a$$
Case 2: $b=-2$
$$d=\dfrac{2}{b} = -1 \\ 6b+d+1 = 6(-2)+(-1)+1=-12 = a$$
So, the options are $(a,b) \in \{(-7,-1),(-12,-2)\}$.
A: Multiplying out to try to get to $6x^4-7x^3+ax^2+3x+2$ gives
\begin{eqnarray*}
(x^2-x+b)(6x^2-x+c) = 6x^4-7x^3+ax^2+3x+2.
\end{eqnarray*}
Equating coefficients 
\begin{eqnarray*}
cb=2 \\
c+b=-3 \\
6b+1+c =a.
\end{eqnarray*}
Should be easy from here ?

 $(a,b)=(-7,-1)$ or $(a,b)=(-12,-2)$

A: 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, $(a,b)\in\{(-7,-1),(-12,-2)\}$.
A: Writing $P=Q\times P_2$ where $P_2$ represents a quadratic. Hence $P=(Q-b)\cdot P_2+b\cdot P_2$
$P=(x^2-x)\cdot P_2+b\cdot P_2$
Terms with powers higher than 2 can only come from the first factor. Comparing with coefficients of $x^4,x^3$ gives $P_2=6x^2-x+k$ where $k$ is a constant.
$\implies P=(x^2-x)(6x^2-x+k)+6bx^2-bx+bk$.
Comparing the coefficients again for $x^2, x, x^0$ gives
$k+1+6b=a$
$-k-b=3$
$bk=2$
Hence $(a,b)=(-7,-1),(-12,-2)$
A: Performing the Euclidean division of $P(x)$ by $Q(x)$ yields
$$P(x)=(6x^2-x+(a-b-1)(x^2-x+b)+(a+2)x-b(a-b-1)+2,$$
whence the conditions
$$a=-2,\qquad b(b+3)+2=0.$$
