# Thread: Irreducible polynomials

1. ## 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....

2. 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 $\mathbb{R}[x]$

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.