Synthetic division by $ax^2+bx+c$. I know that synthetic division can be used in order to find quotient $q(x)$ and remainder $r(x)$ of a polynomial $p(x)$ when it is divided by some linear polynomial like $x-c$. Now, does exist some procedure (another than long division) in order to find $q(x)$ and $r(x)$ when the divisor is $ax^2+bx+c$? Suppose $b^2-4ac<0$.
 A: I suppose you could use comparing coefficients.
$$ \begin{align*} p(x) & = q(x)(ax^{2}+bx+c) + r(x) \\
\alpha_{n}x^{n} + \ldots + \alpha_{1}x + \alpha_{0} & = (\beta_{n-2}x^{n-2} + \ldots +\beta_{1}x+\beta)(ax^{2}+bx+c) + (mx+k) \end{align*}
$$
Now if you multiplied out and simplified all of the right hand side and equated the coefficient of $x^{j}$ for each $j,$ you would have $n$ equations and $n$ unknowns, so you can find out $\beta_{j}, m$ and $k.$
Note that the degree of $r(x)$ is provided for us by the remainder theorem.
A: Assume your divisor is $x-d$ rather than $x-c$ since $c$ is already used in the formation of the original polynomial. You can write the dividend $y$ as:
$y = a(x-d)^2 + b(x-d) + 2adx - ad^2 + c + bd = a(x-d)^2 + b(x-d) + 2ad(x-d) + 2ad^2 -ad^2 + c + bd = a(x-d)^2  + (b+2ad)(x-d) + ad^2 + bd + c = (x-d)(a(x-d) + b+2ad) + ad^2+bd+c = (x-d)(ax + ad + b) + ad^2+bd+c \Rightarrow q(x) = ax+ad+b, r(x) = ad^2+bd+c$.
A: For $ax^2 + bx +c$ with $b^2 - 4ac $>0 find the roots of the polynomial and apply the syntectic division twice in a row .
