# Homework Help: Irreducibility of a polynomial by eisenstein and substitution

1. Nov 23, 2009

1. The problem statement, all variables and given/known data
Show $$16x^4 = 8x^3 - 16x^2 - 8x + 1$$ is irreducible.

2. Relevant equations

Eisenstein's criteria, if there is n s.t. n does not divide the leading coefficient, divides all the other coefficients, and n^2 does not divide the last coefficient then the polynomial is irreducible (over the rationals)

3. The attempt at a solution
I want to say that consider $$p'(x) = p(\frac{1}{2}x) = x^4 + x^3 - 4x^2 - 4x + 1$$ is irreducible by eisenstein if we use the standard trick of substituting x+1 -> x, then we get, $$x^4+5x^3+5x^2-5x-5$$ where eisenstein is immediate. What I don't know is that if I can then say that since p' is irreducible, then p is.

2. Nov 24, 2009

### HallsofIvy

This should be [tex]16x^2+ 8x^4- 16x^2- 8x+ 1[/itex]

p(x) is reducible if and only if p(x)= a(x)b(x) for polynomials a and b of lower degree than p. In that case p(x/2)= a(x/2)b(x/2) so, yes, if p(x) is reducible so is p(x/2). And, then, if p(x/2) is irreducible, so is p(x).

3. Nov 28, 2009