Factorisation of cyclotomic polynomials

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x^16 + 1 is irreducible over the rationals, correct?...


...also I am required to factorise the following polynomials

2) i) x^5 + 3x^4 + 2x^3 + x^2 -7

ii) x^5 + 10x^4 + 13x^3 -25x^2 -68x -60


now I would usually approach this using the factor theorem to find a factor and then divide by this factor and continue, however in the question I am told as a hint that I am to try substituting x-> x+h h=+-1, +-2, why all this is necessary i canot think?...
 
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"x^16 + 1 is irreducible over the rationals, correct?..."

yup..

As for your hint, that is usually used to alter the roots of an equation. eg of the roots of a polynomial of degree 3 in x, is alpha, beta and gamma, subbing in x+1 will get you a polynomial that has roots alpha-1, beta -1 , gamma -1 etc etc.

Maybe that's a hint your ment to use creatively?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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