While attempting to divide a quartic by a quadratic factor to find the other factors of the given quartic, I can't help feeling I "invented" a way of dividing polynomials.

Suppose you have a quartic polynomial $A$ and a quadratic $B$ where you know that $B$ is a factor of $A$.

You will have:

$$\frac AB=\frac{x^4+ax^3+bx^2+cx+d}{x^2+px+q}=C?$$

Now, performing synthetic division is not possible and performing long division is likely to produce mistakes, so I sort of cheated?

I know $B$ goes into $A$ perfectly, which means polynomial $C$ is a quadratic in the form:


Now, I did a sort of cheatery.

I substituted $x=0$ everywhere to obtain:

$$\frac dq=u$$

Then I substituted $x=1$ to get:

$$\frac{1+a+b+c+d}{1+p+q}=1+t+u=1+t+\frac dq$$

And so, you can easily solve for $t,u$ through my method.

My question is whether or not this is a valid way of performing polynomial division, given that the divisor is a factor of the dividend. (or the other way around if I mixed up divisor and dividend.)

  • $\begingroup$ Sure. And we can easily deal with the special cases where the above calculation involves division by $0$. $\endgroup$ – André Nicolas Feb 11 '16 at 23:38
  • $\begingroup$ @AndréNicolas Hold up, you mean to tell me that I can deal with the division by $0$ in this problem? Could you please elaborate, as it is of some importance to something I am doing. $\endgroup$ – Simply Beautiful Art Feb 11 '16 at 23:43
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    $\begingroup$ There is no problem with your method whatsoever. Alternatively, one can multiply $B$ and $C$ and then equate terms with the same power of $x$ in order to compute $t$ and $u$. $\endgroup$ – Pavel Feb 11 '16 at 23:44
  • $\begingroup$ @SimpleArt: There are ad hoc methods. If the polynomials are both divisible by $x$, cancel, we are down to a simpler problem. If they both have $x=1$ as a root, use $x=-1$. If that doesn't work either, the quotient is $x^2-1$. But this may be irrelevant to your other problem. $\endgroup$ – André Nicolas Feb 11 '16 at 23:51
  • $\begingroup$ @AndréNicolas No, what you described is exactly what I did for a much harder problem related to this.... $\endgroup$ – Simply Beautiful Art Feb 11 '16 at 23:58

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