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October 20th, 2007, 12:31 AM  #1 
Newbie Joined: Aug 2007 Posts: 28 Thanks: 0  finding the roots
i have an equation... 2x^6  3x^5 +5x^4 + 6x^3 27x + 81 = 0 (sqrt of 2)  i is given to be a root now this logically implies that the conjugate is also a root i.e. (sqrt of 2) + i...proceeding from here do i use the relations between the roots and the coefficients..say sum of roots...is 3/2 product of roots is 27/2 and so on..but this means i have to use the other relations like sum taken 5 at a time and so on to get all the roots....which is rigorous..is there any other simpler way of doing this??? 
October 20th, 2007, 12:39 PM  #2 
Senior Member Joined: Sep 2007 From: USA Posts: 349 Thanks: 67 Math Focus: Calculus 
Since you know 2 roots, you can use synthetic division for your polynomial divided by (x^2  2âˆš(2)x + 3). This reduces to a 4th degree polynomial, but the coefficients are no longer integers. One way to solve this is to use a quartic formula to solve for the remaining roots.

October 20th, 2007, 07:14 PM  #3 
Newbie Joined: Aug 2007 Posts: 28 Thanks: 0 
well is there any other way??i feel thre is..i am just not able to catch onto it...

October 20th, 2007, 07:19 PM  #4 
Senior Member Joined: Sep 2007 From: USA Posts: 349 Thanks: 67 Math Focus: Calculus 
Well, you could use the fact that two complex roots and their conjugates are the remaining solutions somehow.

October 20th, 2007, 08:40 PM  #5 
Global Moderator Joined: Dec 2006 Posts: 18,422 Thanks: 1462 
(x²  2?(2)x + 3)(x² + 2?(2)x + 3) = x^4  2x² + 9 2x^6  3x^5 + 5x^4 + 6x³  27x + 81 = (2x²  3x + 9)(x^4  2x² + 9) You can easily complete the factorization now. 
October 20th, 2007, 09:07 PM  #6 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 937 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms 
It's a quadratic in x².

October 20th, 2007, 11:49 PM  #7 
Newbie Joined: Aug 2007 Posts: 28 Thanks: 0 
Well thanks skipjack... but please explain why you did this (x²  2?(2)x + 3)(x² + 2?(2)x + 3) as only the first term is the factor... as gleaned by the fact that ?2  i is a root ... how did you find out that ?2  i is also a root?

October 21st, 2007, 08:29 AM  #8 
Global Moderator Joined: Dec 2006 Posts: 18,422 Thanks: 1462 
If division by one of the factors works, at precisely what stage could it fail for the other?


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2x^63x^5 5x^4 6x^327x 81,2x^63x^5 5x^4 6x^327x 81=0 solve equation,solve 2X^63x^5 5x^4 6x^327x 81=0,one root of equation 2x^63x^5 5x^4 6x^327x 81=0 is √2 i find the remaining roots?
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