Consider this simple 5th order polynomial:

(*   (1+x+x^2)(1-x+x^3)   *)
 root=FindInstance[pol==0, x]
(*   -1/2 -I/2 Sqrt[3]   *)

Suppose we would like to see the 4th order polynomial with this root removed. This does not immediately seem to be easy... For instance:


are both just returning the fraction of a 5th order polynomial divided by a 1st order one... How to see the 4th order result explicitly?

(PS: I know it could become a very messy expression, but then I would try one of the other roots. I just need to see the result in order to judge that.)


2 Answers 2


PolynomialQuotient is enough. I just used PolynomialQuotientRemainder to see that remainder is indeed 0.

PolynomialQuotientRemainder[x^5 + x^4 + 1, x - (-1/2 - I/2 Sqrt[3]), x]

$$\left\{x^4+\left(\frac{1}{2}-\frac{i \sqrt{3}}{2}\right) x^3-x^2+\left(\frac{1}{2}+\frac{i \sqrt{3}}{2}\right) x-\frac{i \sqrt{3}}{2}+\frac{1}{2},0\right\}$$

  • $\begingroup$ Thanks, that solves it... $\endgroup$ Commented Apr 27 at 9:42

FindInstance returns a rule, not a number. Therefore you need:

root = x /. FindInstance[pol == 0, x][[1]]

1/2 (-1 - I Sqrt[3])

To divide the polynomial by the root:

red = PolynomialQuotient[pol, (x - root), x]

enter image description here

Further, "Factor" factors a polynomial over the integers, not the Complexes. Therefore, to get all the roots of "red":

Solve[red == 0, x] // N

{{x -> -0.5 + 0.866025 I}, {x -> -1.32472}, {x -> 
   0.662359 - 0.56228 I}, {x -> 0.662359 + 0.56228 I}}

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