Re: Check a polynomial for (n-1) equal roots?

In sci.math, Gerry
<GerryMrt@xxxxxxxxx>
wrote
on 22 Mar 2007 04:22:52 -0700
<1174562572.443280.196630@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>:
Hi, which is the best way to determine that a polynomial in one
variable with real coefficients
of degree n has (n-1) equal roots?

Is there a general methode to verify for (n-m) equal roots?

Any comments are welcome

Gerry


If a polygon has (n-1) equal roots, n > 2 (n = 2 is a special case
which isn't really worth pursuing) then it is of the form

(x+p)^(n-1) * (x+q)

for some p and q. This expands into

a_n * x^n + a_{n-1} * x^{n-1} + ... + a_1 * x + a_0

where
a_n = 1 = (n-1)!/(0!(n-1)!)
a_{n-1} = q + p*(n-1)!/(1!(n-2)!)
a_{n-2} = q * p * (n-1)!/(1!(n-2)!) + p^2 * (n-1)!/(2!(n-3)!)
a_{n-3} = q * p^2 * (n-1)!/(2!(n-3)!) + p^3 * (n-1)!/(3!(n-4)!)

At this point I have three equations in two unknowns; if one can
still solve them one has a fighting chance. One can then keep
going with these equations until one gets to

a_0 = p^(n-1) * q

If one can "run the gauntlet" (and if I've done this correctly!),
the equation is of your stipulated form.

As an example, take the equation

x^5 - 23/3*x^4 + 64/3*x^3 - 24*x^2 + 16/3*x + 16/3

where n = 5. The first two equations then become

-23/3 = q + 4*p
q = -23/3 - 4*p

64/3 = 4*q*p + 6*p^2
= 4*(-23/3-4*p)*p + 6*p^2

or

10*p^2 + 92/3*p + 64/3 = 0

The roots of this equation are p = -16/15 and p = -2, and
q = -17/5 in the first case, 1/3 in the second.
We now look at the third equation:

-24 = 6*q*p^2 + 4*p^3

6*(-17/5)*(-16/15)^2 + 4*(-16/15)^3 = -18944/675, which
clearly doesn't work; however,
6*(1/3)*(-2)^2 + 4*(-2)^3 = -24

works absolutely fine.

16/3 = 4*q*p^3 + p^4

is also true, and

16/3 = p^4*q

so the given equation is of this form:

x^5 - 23/3*x^4 + 64/3*x^3 - 24*x^2 + 16/3*x + 16/3 = (x-2)^4*(x+1/3)

(Yes, I cheated. :-) But in any event the procedure should work.
Sometimes the straightforward ways are the simplest.)

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