The math behide this solution is quite simple:
if a polynomial equation f(x)=0 has roots, let's say x1,x2,……xn,then
f(x) can be converted to another format as
(x-x1)*(x-x2)*(x-x3)*…...*(x-xn) = 0
So what we do here is use goal-seek to find it's first root(x1)
then use f(x) / (x-x1) as a new equation, easy to say the new one is
(x-x2)*(x-x3)*…...*(x-xn) = 0
repeat the procedure above,we can get all the roots one by one.


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Pan
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I know know if excel is the best way to go here....
There's a program called Maple that would be of use to you...
www.maplesoft.com

However, if there is an excel solution, I'm sure someone on here can figure
it out.

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"Pan" wrote:


The math behide this solution is quite simple:
if a polynomial equation f(x)=0 has roots, let's say x1,x2,€¦€¦xn,then
f(x) can be converted to another format as
(x-x1)*(x-x2)*(x-x3)*€¦...*(x-xn) = 0
So what we do here is use goal-seek to find it's first root(x1)
then use f(x) / (x-x1) as a new equation, easy to say the new one is
(x-x2)*(x-x3)*€¦...*(x-xn) = 0
repeat the procedure above,we can get all the roots one by one.


+-------------------------------------------------------------------+
|Filename: Find all the roots of High-Order Polynomial Equation by Dragon.Pan.zip|
|Download: http://www.excelforum.com/attachment.php?postid=3956 |
+-------------------------------------------------------------------+

--
Pan
------------------------------------------------------------------------
Pan's Profile: http://www.excelforum.com/member.php...o&userid=28144
View this thread: http://www.excelforum.com/showthread...hreadid=480008