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We know about polynomials. A polynomial equation of $n^{th}$ degree would be:

$$a_1x^n + a_2x^{n-1} + a_3x^{n-3} + ... + a_n = 0$$

I have also been told that it's quite difficult to solve polynomial equations above degree 2.

But is it possible with some manipulation of the coefficients, to convert that polynomial, into a system of $n$ linear equations, such that solving the system would give me the all the roots of the polynomial equations?

Pritt Balagopal
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    The short answer is no. General polynomial so of degree greater than $5$ cannot be solved in radicals. However radicals are beyond the reach of linear algebra. – Michael Burr May 19 '17 at 02:24
  • @MichaelBurr Why not? – Pritt Balagopal May 19 '17 at 02:24
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    Because polynomials are not linear for $n > 1$ – Manuel Guillen May 19 '17 at 02:27
  • @ManuelGuillen Polynomials arent linear, Im quite aware of that. I want to find roots of it, so I'm looking for transformation to $n$ linear equations instead. – Pritt Balagopal May 19 '17 at 02:45
  • You can't break down something such as $x^2 = 1$ into a system of linear equations, because if you tried, the coefficients would not be constant. If thats not enough proof, @MichaelBurr made the valid point that its been proven that it is impossible to derive a formula for the roots of a polynomial of order greater than or equal to 5. – Manuel Guillen May 19 '17 at 03:00
  • @ManuelGuillen of order greater than four. – law-of-fives May 19 '17 at 03:01
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    @Manuel, there's a difference between "cannot be solved in radicals" and "impossible to derive a formula" ; the first is correct, but the second one isn't. – J. M. ain't a mathematician May 19 '17 at 03:55
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    Pritt: you might want to see this. – J. M. ain't a mathematician May 19 '17 at 03:56
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    If your question is if there exists a linear system with the roots as solutions, the answer is there exists infinitely many (possibly with complex coefficients). However IMHO there is no way to get any of those linear systems without first solving the roots in the first place, or of equivalent complexity! So the approach isn't helpful. – Macavity May 19 '17 at 05:35
  • @Macavity Thanks for your insight. But how can you confidently state that there's no way? Couldn't it be that a way hasn't been discovered yet? – Pritt Balagopal Jun 06 '17 at 06:49
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    @PrittBalagopal That's why I mentioned it's my opinion. Nothing stops future possibilities, but at this moment that's in the realm of conjecture. – Macavity Jun 06 '17 at 07:14

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