Sorry,I could say clearly and have some carelessness.

n=2,no parameter.
(a1+b1x)(a2+b2x)=-b1(-a1b2/b1-b2x)(a2+b2x)/b2.We have
x=(-a2b1-a1b2)/2b1b2.
n=3,1parameter. c3
(a1+b1x)(a2+b2x)(a3+b3x)=-b1((-a1(b2+c3b3)/b1-(b2+c3b3)x)(a2+b2x)(a3c3+b3c3x)/b2c3(b2+c3b3)
We have,a2+b2x=1/3(a3c3+a2-a1(b2+c3b3)/b1)
a3c3+b3c3x=1/3(a3c3+a2-a1(b2+c3b3)/b1)
We can get x.
For n,n-2 parameters,c3,..cn.
(a1+b1x)(a2+b2x)...(an+bnx)=-b1((-a1(b2+c3b3+...cnbn)/b1-(b2+c3b3...cnbn)x)(a2+b2x)...(ancn+bncnx)/b2c3...cn(b2+c3b3+...cnbn)
We have,a2+b2x=1/n(a2+a3c3+...ancn-a1(b2+c3b3...cnbn)/b1)
a3c3+b3c3x=1/n(a2+a3c3+...ancn-a1(b2+c3b3...cnbn)/b1)...
ancn+bncnx=1/n(a2+a3c3+...ancn-a1(b2+c3b3...cnbn)/b1)...
We can get x.
when we construct fisrt factor,we lost one independent Equation.

所有跟帖： 

• I really appreciate your effort. -皆兄弟也- ♂ (0 bytes) () 09/14/2010 postreply 10:13:56

• For n,it seems (n-1)th equition should be delt with. -皆兄弟也- ♂ (0 bytes) () 09/14/2010 postreply 10:22:32

• really wonderful. there are no (n-1)th equition . -皆兄弟也- ♂ (259 bytes) () 09/14/2010 postreply 10:43:52