# How to remove duplicate roots from a polynomial?

Given a polynomial equation (with real coefficients of any degree with any number of repeating roots), let say

$x^5 + 6x^4 - 18x^3 - 10x^2 + 45x - 24 = 0$, ... (A)

it can be written as

$(x-1)^2 (x+8) (x^2-3) = 0$.

As you can see, it has a repeating root $x = 1$.

Forming a new polynomial by removing the duplicate root (i.e, taking repeating roots only once), we will get

$(x-1) (x+8) (x^2-3) = 0$.

Then expanding it, we get

$x^4+7 x^3-11 x^2-21 x+24 = 0$. ... (B)

Is it possible to get a polynomial having all of the roots of the given polynomial but taking the repeating roots only once, i.e, getting (B) from (A), directly without finding any roots? If possible, how to do the same?

• How does one know that there is a repeated root without knowledge of any root? Apr 10 '15 at 17:11
• It is a polynomial condition on the coefficients of the polynomial, also known as $Res(f,f')=0$. @Dr.MV cfr Resultant
– user228113
Apr 10 '15 at 17:16
• @Dr.MV : I don't know. This question came to my mind but I could not think of a way to do it. I am expecting that the algorithm/solution should return the same polynomial if it has no repeating roots. Apr 10 '15 at 17:20
• The squarefree part is $\, f/\gcd(f,f')\,$ over a field of characteristic 0 Apr 10 '15 at 17:25
• @G.Sassatelli Does that tell one whether the original polynomial of interest has multiple roots? And how does one proceed to find the "resultant" without any knowledge of any root? Apr 10 '15 at 17:26

$${\rm rad}(f(x))\, =\, \frac{f(x)}{\gcd(f(x),f'(x))}\qquad$$
This works because taking the dervative decrements the multiplicity of each prime factor, thus taking the above quotient has the effect of cancelling out repeated factors from $\,f.\,$ (Beware that this need not work over fields of characteristic $\neq 0).$