3
$\begingroup$

Consider this simple 5th order polynomial:

 pol=x^5+x^4+1
 Factor[pol]
(*   (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:

FullSimplify[pol/(x-root)]
Factor[pol/(x-root)]

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.)

$\endgroup$

2 Answers 2

3
$\begingroup$

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]

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

$\endgroup$
1
  • $\begingroup$ Thanks, that solves it... $\endgroup$ Commented Apr 27 at 9:42
3
$\begingroup$

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}}
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.