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 $b^2-4ac<0$.

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    $\begingroup$ You have detailed explanations and examples on wikipedia. $\endgroup$
    – Bernard
    Dec 28, 2014 at 23:33
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    $\begingroup$ There's an example here on MathWorld -- but its not really clear to me that it's any different from ordinary polynomial division, except with a different convention for where on the paper to write down the intermediate results. $\endgroup$ Dec 28, 2014 at 23:44

3 Answers 3


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.


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 .

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    $\begingroup$ I am interested in the case $b^2-4ac<0$. Thanks. $\endgroup$ Dec 29, 2014 at 4:06
  • $\begingroup$ @ÁngelMarioGallegos That makes no difference. $\endgroup$ May 19 at 6:06

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:

$y = a(x-d)^2 + b(x-d) + 2adx - ad^2 + c + bd = a(x-d)^2 + b(x-d) + 2ad(x-d) + 2ad^2 -ad^2 + c + bd = a(x-d)^2 + (b+2ad)(x-d) + ad^2 + bd + c = (x-d)(a(x-d) + b+2ad) + ad^2+bd+c = (x-d)(ax + ad + b) + ad^2+bd+c \Rightarrow q(x) = ax+ad+b, r(x) = ad^2+bd+c$.

  • $\begingroup$ The question asks when the 'divisor' is $ax^{2}+bx+c.$ Haven't you assumed the 'dividend' is $ax^{2} + bx + c$ and the 'divisor' $x-d$? $\endgroup$
    – Shai
    Dec 28, 2014 at 23:48
  • $\begingroup$ I want to know the quotient and the remainder of $p(x)\div (\mathbf{ax^2+bx+c})$ where $b^2-4ac<0$ and $p(x)$ is any polynomial. $\endgroup$ Dec 29, 2014 at 4:25
  • $\begingroup$ Take as an example $p(x)=3x^4+4x^2-x+1$ divided by $x^2+2x+5$. I have no idea how what you have posted can be used in order to find the quotient and remainder for this division. $\endgroup$ Dec 29, 2014 at 6:34
  • $\begingroup$ I am asking for another way (other than long division). Thank you, anyway. $\endgroup$ Dec 29, 2014 at 6:43
  • $\begingroup$ @ah-huh-moment. Thanks anyway. $\endgroup$ Dec 29, 2014 at 6:58

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