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October 19th, 2007, 11:31 PM   #1
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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 co-efficients..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???
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October 20th, 2007, 11:39 AM   #2
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Since you know 2 roots, you can use synthetic division for your polynomial divided by (x^2 - 2√(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.
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October 20th, 2007, 06:14 PM   #3
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well is there any other way??i feel thre is..i am just not able to catch onto it...
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October 20th, 2007, 06:19 PM   #4
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Well, you could use the fact that two complex roots and their conjugates are the remaining solutions somehow.
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October 20th, 2007, 07:40 PM   #5
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(x - 2?(2)x + 3)(x + 2?(2)x + 3) = x^4 - 2x + 9

2x^6 - 3x^5 + 5x^4 + 6x - 27x + 81 = (2x - 3x + 9)(x^4 - 2x + 9)

You can easily complete the factorization now.
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October 20th, 2007, 08:07 PM   #6
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It's a quadratic in x.
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October 20th, 2007, 10:49 PM   #7
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Well thanks skipjack... but please explain why you did this (x - 2?(2)x + 3)(x + 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?
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October 21st, 2007, 07:29 AM   #8
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If division by one of the factors works, at precisely what stage could it fail for the other?
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