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 $Z$
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 \ \ . $