# Irreducible polynomials

• Apr 5th 2006, 08:51 PM
suedenation
Irreducible polynomials
Hi, does anyone know how to do the following questions?

In each of the following cases, find all roots of f(x) in F and factor f(x) as the product of irreducible polynomials in F[x].

(a) F=R, f(x) = (x^6)-1 in R[x]
(b) F=Z3, f(x) = (x^5)+x+1 in Z3[x]
(c) F=Z5, f(x) = (x^3)+2x+3 in Z5[x]

Thanks very much guys.... :)
• Apr 6th 2006, 06:31 AM
ThePerfectHacker
Quote:

Originally Posted by suedenation
Hi, does anyone know how to do the following questions?

In each of the following cases, find all roots of f(x) in F and factor f(x) as the product of irreducible polynomials in F[x].

(a) F=R, f(x) = (x^6)-1 in R[x]

You have, $x^6-1=0$ thus find all real values such as $x^6=1$. This only happens for $x=-1,1$ (note the other 4 roots are complex we are only concerned with real). Thus, $x+1$ and $x-1$ are factors. Dividing $x^6-1$ by $(x-1)(x+1)=x^2-1$ we get,
$x^4+x^2+1$. Thus, the polynomial $x^6-1$ fractors as
$(x-1)(x+1)(x^4+x^2+1)$ in $\mathbb{R}[x]$

Quote:

Originally Posted by suedenation
(b) F=Z3, f(x) = (x^5)+x+1 in Z3[x]

To factor $x^5+x+1$ in $\mathbb{Z}_3[x]$ we simply see what are its zero's. This is easy to check because they are only 3 elements in this field. Checking we see that 1 is a zero because, $1^3+1+1=0$ under addition and multiplication modulo 3. Thus, $x-1$ is a factor. Dividing these polynomials we get, (working modulo 3).
$x^4+x^3+x^2+x+2$
Checking the zero's of this polynomial we find that 1 is it zero. Thus, $x-1$ is a factor. Dividing these polynomials we get, (working modulo 3).
$x^3+2x+1$, notice this polynomial has no zero and because it has a degree of 3 it is irreducible (this is a theorem about the irreducibility of 2 and 3 degree polynomials). Thus, it factors as,
$(x-1)(x-1)(x^3+2x+1)$ or another way of writing this, $(x+1)(x+1)(x^3+2x+1$

I do part (c) latter unless you understand and I could stop.