If I told you a certain polynomial P(x) had a double root (only one!), how could you go about finding it, WITHOUT trying to find every root? Also, the EXACT value of the root is sought; not an approximation.

NB. Roots may be any kind --real or complex-- but they are all different, with multiplicity "1", except for one that has multiplicity "2".

r is a multiple root of P(x) if and only if P(r)=P'(r)=0.

Dividing P(x) by P'(x), we may write P(x)=P'(x)Q(x)+R(x) where the degree of R(x) will be 1 since the degree of P'(x) is one less than that of P(x). But since R(x)= P(x)-P'(x)Q(x), R(r)=0, and so r is a root of the equation R(x)=0. We can then try working with R(x) which has degree at least 2 less than the degree of P(x).

There is a slight problem if P(x) has only the one (double) root r, as then the remainder R(x) will be 0.  But this can be  readily coped with.  This method with the similar minor problem  also applies when r is a multiple root with multiplicity greater than 2.

Edited on May 8, 2006, 4:04 pm

Posted by Richard on 2006-05-08 13:08:49