# How to reduce the multiplicity of existing real roots without introducing new real roots?

Given a monic polyomial $P(x)=x^d+r_{d-1}x^{d-1}+\cdots+a_1r+a_0\in\mathbb{R}[x]$ is there a way to manipulate the coefficients of $P$ in an algebraic way such that the new polynomial has exactly as many simple real roots as $P$ has pairs of twofold roots? For example if $d=9$ and

$$P(x)=x^9-15x^8+95x^7-333x^6+719x^5-1029x^4+1057x^3-819x^2+432x-108=(x-1)^2(x-2)^2(x-3)^3(x^2+1)$$

then the transformed polynomial (the coefficients of which should be polynomial expressions in $r_0,\ldots,r_8$) should have exactly two simple real root corresponding to the twofold roots $1$ and $2$ of $P$ (it can have an arbitrary amount of additional multiple real roots or complex roots though).

One trick that almost does it is $Q(x)=P(x)^2+P'(x)^2$. This polynomial has no simple real roots and its twofold real roots are exactly the twofold real roots of $P$. If there was a way to reduce the multiplicity of the twofold real roots without introducing new real roots this would be the solution.

Another idea is to take $\gcd(P(x),P'(x))$ which has exactly as many simple real roots as $P$ has twofold roots. But one cannot compute the gcd in an alebraic way. One needs to know $P$ in order to use the Euclidean algorithm.

• Notice that computing your Q also requires "knowing P", whatever it is youmean by that. – Mariano Suárez-Álvarez Jul 26 '15 at 3:01

## 2 Answers

Yun's Algorithm will give you a series of polynomials, each corresponding to factors of that multiplicity in the original polynomial. $p_1$ will have all the roots of multiplicity $1$, $p_2$ will have all the roots of multiplicity $2$, etc.

It requires you to take derivatives, perform gcd, and do basic arithmetic with polynomials. It's deterministic. It doesn't actually give you the roots though.

• Thank you for the comment but as I said I can't use the gcd, see my comment to the answer below. – Mario Jul 26 '15 at 4:56

You have to know $P$ in order to do anything with it. So you can compute $gcd(P, P')$ in an entirely algebraic way.

In other words, I don't know what you mean when you write "One needs to know P in order to use the Euclidean algorithm."

• What I mean is that I need an expression involving a general P which only uses the coefficients of P in an algebraic (actually polynomial) way.If $P(x)=x^d+r_{d-1}x^{d-1}+\cdots+a_1r+a_0\in\mathbb{R}[x]$ then $(P(x)^2+P'(x)^2$ can be computed but $\operatorname{gcd}(P,P')$ cannot because I don't know how many steps to perform in the Euclidean algorithm without knowing the coefficients, i.e. I can't do it for a general $P$. I need one and only one expression and no case studies. – Mario Jul 26 '15 at 4:54