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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?

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    $\begingroup$ 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. $\endgroup$ – Michael Burr May 19 '17 at 2:24
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    $\begingroup$ Because polynomials are not linear for $n > 1$ $\endgroup$ – Manuel Guillen May 19 '17 at 2:27
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    $\begingroup$ @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. $\endgroup$ – J. M. is a poor mathematician May 19 '17 at 3:55
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    $\begingroup$ Pritt: you might want to see this. $\endgroup$ – J. M. is a poor mathematician May 19 '17 at 3:56
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    $\begingroup$ 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. $\endgroup$ – Macavity May 19 '17 at 5:35

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