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Find roots of polynomial of degree 7

Solve
x^7 -7x^6 -21x^5 + 35x^4 + 35x^3-21x^2 -7x +1 =0

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Reply 1
Avatar for Notnek
Notnek
21
Original post by B_9710
Solve
x^7 -7x^6 -21x^5 + 35x^4 + 35x^3-21x^2 -7x +1 =0

Find all of the roots? One root isn't too hard to spot.

Are you sure you've got all the signs correct? And where did you get the question from?
Reply 2
Avatar for TeeEm
TeeEm
19
Original post by B_9710
Solve
x^7 -7x^6 -21x^5 + 35x^4 + 35x^3-21x^2 -7x +1 =0



look for an obvious solution

divide it out/factor out

divide the equation by x3

then there is a standard substitution that will reduce it to a cubic ...

or preferably think complex numbers

EDITED
(edited 8 years ago)
Reply 3
Avatar for Notnek
Notnek
21
Original post by TeeEm
divide the equation by x4

then there is a standard substitution that will reduce it to a cubic ...

or preferably think complex numbers

Ah yes.
Reply 4
Avatar for TeeEm
TeeEm
19
Original post by notnek
Ah yes.


it looks like a binomial expansion but the pattern ++--++-- etc indicates powers of i

I leave it with you as I did not see your initial comment to the OP and I am busy doing work
Reply 5
Avatar for Waliid
Waliid
8
X^6-8x^5-13x^4+48x^3-13x^2-8x+1
X7=8.87525


Posted from TSR Mobile
Reply 6
Avatar for Notnek
Notnek
21
Original post by TeeEm
it looks like a binomial expansion but the pattern ++--++-- etc indicates powers of i

I leave it with you as I did not see your initial comment to the OP and I am busy doing work

I'm going to bed :smile:

Off subject, I'm never allowed to rep you for some reason. I owe you a lot of rep!
Reply 7
Avatar for TeeEm
TeeEm
19
Original post by notnek
I'm going to bed :smile:

Off subject, I'm never allowed to rep you for some reason. I owe you a lot of rep!


no worries.

(I can always rep because I tend to be to generous with it, as I rep at least 20 posts /regular helpers every day... I am not quite sure what is the number but you cannot rep the same person again until you rep others(?) exactly how many I do not know)
Reply 8
Avatar for B_9710
B_9710
OP
18
I do have the answers. With the general solution being x=tan(nπ/7 + π/28)
I was just seeing the different ways that people suggest to go about this question
Reply 9
Avatar for TeeEm
TeeEm
19
Original post by B_9710
I do have the answers. With the general solution being x=tan(nπ/7 + π/28)
I was just seeing the different ways that people suggest to go about this question


see my edited post then

I will write a full solution and post it later (using complex numbers)
(edited 8 years ago)
Original post by B_9710
I do have the answers. With the general solution being x=tan(nπ/7 + π/28)
I was just seeing the different ways that people suggest to go about this question
Given the general solution it's pretty clear how to solve this in about 4 lines, but I'm not sure I'd have spotted it without some hint that trig-type solutions were going to come in.

Edit: (actually, I think I would have done, but I'd have basically gone about 8 lines into doing it one way and then started again...)
(edited 8 years ago)
Reply 11
Avatar for TeeEm
TeeEm
19
Original post by B_9710
I do have the answers. With the general solution being x=tan(nπ/7 + π/28)
I was just seeing the different ways that people suggest to go about this question


As you posted this as a challenge to us and not actually not actually requiring help, here is my solution to a very standard type equation, which I hope other students might find of interest.

(your question will now found itself condemned into the abyss of my own resources ...)
(edited 8 years ago)
IMG_0001.jpg386.4KB
IMG_0002.jpg452.3KB
Reply 12
Avatar for davros
davros
16
Original post by DFranklin
Given the general solution it's pretty clear how to solve this in about 4 lines, but I'm not sure I'd have spotted it without some hint that trig-type solutions were going to come in.

Edit: (actually, I think I would have done, but I'd have basically gone about 8 lines into doing it one way and then started again...)


As TeeEm spotted earlier, the coefficients are basically row 7 of Pascal's triangle BUT the pattern of signs suggests there's going to have to be some 'fiddling' to get it to work out.
Reply 13
Avatar for B_9710
B_9710
OP
18
Original post by davros
As TeeEm spotted earlier, the coefficients are basically row 7 of Pascal's triangle BUT the pattern of signs suggests there's going to have to be some 'fiddling' to get it to work out.


That's the point. If the coefficients were randomly chosen and were not patterns of the Pascal's triangle then the way to find the roots would be completely different. I just posted this to see how others recognise this fact and then how people go about solving it.
Reply 14
Avatar for davros
davros
16
Original post by B_9710
That's the point. If the coefficients were randomly chosen and were not patterns of the Pascal's triangle then the way to find the roots would be completely different. I just posted this to see how others recognise this fact and then how people go about solving it.


If the coefficients were 'randomly chosen' then the way to find the roots exactly would be non-existent because degree-7 polynomials aren't soluble by radicals! Of course, you could use numerical methods to approximate the roots to any required degree of accuracy, but then you don't have a particularly attractive problem any more :smile:
Original post by TeeEm
As you posted this as a challenge to us and not actually not actually requiring help, here is my solution to a very standard type equation, which I hope other students might find of interest.

(your question will now found itself condemned into the abyss of my own resources ...)


You have beautiful handwriting.
Reply 16
Avatar for TeeEm
TeeEm
19
Original post by Mihael_Keehl
You have beautiful handwriting.


I wish some of my students could see this comment as most complain about it, and in particular that I write in capitals.
(edited 8 years ago)
Original post by TeeEm
I wish some of my students could see as most complain about it, and in articular that I write in capitals.


If x = 1,-1 is a root to a polynomial that long, then is i or i^2 also a root?

Those students tho
Reply 18
Avatar for TeeEm
TeeEm
19
Original post by Mihael_Keehl
If x = 1,-1 is a root to a polynomial that long, then is i or i^2 also a root?

Those students tho


x=-1 is a solution, x=1 is not

if the coefficients are real then any complex roots will appear in conjugate pairs but this has 7 real solutions
Original post by TeeEm
x=-1 is a solution, x=1 is not

if the coefficients are real then any complex roots will appear in conjugate pairs but this has 7 real solutions


I was talking in general for a polynomial with a high degree, not in this particular case, sorry ifI was vague

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