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

Quote:

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

To factor \$\displaystyle x^5+x+1\$ in \$\displaystyle \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, \$\displaystyle 1^3+1+1=0\$ under addition and multiplication modulo 3. Thus, \$\displaystyle x-1\$ is a factor. Dividing these polynomials we get, (working modulo 3).
\$\displaystyle x^4+x^3+x^2+x+2\$
Checking the zero's of this polynomial we find that 1 is it zero. Thus, \$\displaystyle x-1\$ is a factor. Dividing these polynomials we get, (working modulo 3).
\$\displaystyle 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,
\$\displaystyle (x-1)(x-1)(x^3+2x+1)\$ or another way of writing this, \$\displaystyle (x+1)(x+1)(x^3+2x+1\$

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