Factorisation of cyclotomic polynomials

  • Thread starter Thread starter catcherintherye
  • Start date Start date
  • Tags Tags
    Polynomials
catcherintherye
Messages
47
Reaction score
0
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?...
 
Last edited:
Physics news on Phys.org
"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?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top