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# Find roots of polynomial of degree 7 watch

1. Solve
x^7 -7x^6 -21x^5 + 35x^4 + 35x^3-21x^2 -7x +1 =0
2. (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?
3. (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
4. (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.
5. (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
6. X^6-8x^5-13x^4+48x^3-13x^2-8x+1
X7=8.87525

Posted from TSR Mobile
7. (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

Off subject, I'm never allowed to rep you for some reason. I owe you a lot of rep!
8. (Original post by notnek)
I'm going to bed

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)
9. I do have the answers. With the general solution being x=tan(nπ/7 + π/28)
10. (Original post by B_9710)
I do have the answers. With the general solution being x=tan(nπ/7 + π/28)
see my edited post then

I will write a full solution and post it later (using complex numbers)
11. (Original post by B_9710)
I do have the answers. With the general solution being x=tan(nπ/7 + π/28)
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...)
12. (Original post by B_9710)
I do have the answers. With the general solution being x=tan(nπ/7 + π/28)
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 ...)
Attached Images

13. (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.
14. (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.
15. (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
16. (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.
17. (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.
18. (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
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
20. (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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