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I came across a polynomial equation $$x^6-2x^5-7x^4+8x^3+32x^2-64=0$$

My aim is to find the value of(without any software)$\frac{1}{x-1}$, if $x$ is the real root of the above sixth degree equation.

I will be happy if there is any way or theorem to tell whether a polynomial is reducible or irreducible over $Z$

Ekaveera Gouribhatla
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    Have you heard of the rational root theorem? Also, there are many criterion for irreducibility over Z, e.g. Eisenstein – TheBestMagician Sep 13 '22 at 01:36
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    There are 2 real roots. Why do we find "$x$ is the real root" in the question. – MasB Sep 13 '22 at 01:57
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    Since the Rational Zeroes Theorem tells us that this monic polynomial will only have "candidate" zeroes which are integer powers of 2 , that coefficient $ \ −7 \ $ (all the other coefficients are powers of 2) is going to eliminate all of the candidates. (And the sum of the coefficients of () and (−) is not zero.) So you're out of luck for rational zeroes... –  Sep 13 '22 at 02:40
  • Mathematica seems to be unable to give the closed forms of the roots. – Max0815 Sep 13 '22 at 02:41
  • In general for all polynomial expressions greater then degree 4, there is no general solution in terms of Elementary mathematical functions. Unless this polynomial can be factorized into polynomials of less then or equal to degree 4, then it's real roots cannot be expressed in terms of elementry mathematical functions. This includes +, -, *, / and nth roots – Colonizor48 Sep 13 '22 at 20:37

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To elaborate on my comment, the Rational Zeroes Theorem indicates "candidate" zeroes $ \ \pm 2^m \ \ , \ $ with $ \ 0 \ \le \ m \ \le \ 6 \ \ . \ $ For the coefficients in the polynomial, we would have to test $$ 2^{6m} \ \mp \ 2^{5m+1} \ \pm \ 2^{3m+3} \ + \ 2^{2m+5} \ - \ 2^6 \ - \ (2^3 - 1)·2^{4m} \ $$ to see if any of the sums are zero for the values of $ \ m \ $ in the prescribed interval. (None of them are...)

ADDENDUM (9/13) -- The upper bound on the absolute value of the real zeroes given by Lagrange (as described in the first answer here) is $ \ 2·64^{1/(6-1)} \ \approx \ 2.30 \ \ , \ $ so we could in fact stop our testing with $ \ m = 1 \ \ . $