# Factorisation of a polynomial of degree 5 using limited theorems

I have the polynomials $p(X)=X^{5}+P_4X^{4}+P_3X^{3}+P_2X^{2}+P_1X+P_0 \in\mathbb{Z_2[x]}$
I need to determine all $p(x)$ that can be factored into irreducible polynomials of degree three and two.

The catch is that I cannot use most of the techniques usually described to solve this kind of problem. What would be the steps I need to take in order to solve this problem?

For the record I am using Gallian: contemporary abstract algebra as a book and can only use theorems 17.1 and 17.5(plus corollary 1 and 2) from that specific chapter. For reference I wrote them down below:

17.1: Let $F$ be a field. If $f(x) [ F[x]$ and deg $f(x)$ is 2 or 3, then $f(x)$ is reducible over $F$ if and only if $f(x)$ has a zero in $F$.

17.5: Let $F$ be a field and let $p(x) \in F[x]$. Then $<p(x)>$ is a maximal ideal in $F[x]$ if and only if $p(x)$ is irreducible over $F$.

Corollary 1: Let $F$ be a field and $p(x)$ be an irreducible polynomial over $F$. Then $F[x]/<p(x)>$ is a field.

Corollary 2: Let F be a field and let $p(x), a(x), b(x) \in F[x]$. If $p(x)$ is irreducible over $F$ and $p(x) | a(x)b(x)$, then $p(x) | > a(x)$ or $p(x) | b(x)$.

• If it has an irreducible factor of degree 4, then it also has a linear factor and so has a zero. – Morgan Rodgers Mar 24 '16 at 18:03
• I am not sure I follow. – MSB Mar 24 '16 at 18:07
• A polynomial of degree 5 is either irreducible, splits into irreducible factors of degree 2 and 3, or else has a linear factor (irreducible factors of degree 4 and 1, 3 and 1 and 1, or 2 and 2 and 1). It is easy to check for linear factors by testing if $p(1) = 0$ or $p(0)=0$. – Morgan Rodgers Mar 24 '16 at 19:58

## 1 Answer

Use the no roots criterion to determine all irreducible quadratics and cubics. Quadratics are very simple, there is only $x^2+x+1$. Cubics are a little more work. But not much. The cubic has to have shape $x^3+ax^2+bx+1$ where there is an odd number of $1$'s. Once you have your list of cubics, multiply each by $x^2+x+1$.

• Thanks, this set me on the right path, will check it as an answer when I have solved the problem. – MSB Mar 24 '16 at 18:26
• You are welcome. – André Nicolas Mar 24 '16 at 18:33
• so for quadratics I have: $p(x)= x^{2}+x+1$ and $p(x) = 1$, removing the duplicates at the cubics I also have: $p(x)= x^{3}+x^{2}+1$ and $p(x)= x^{3}+x+1$ So all I need to do is multiply the cubic with the quadratic(that isn't 1) to get all of the polynomials of degree 5 that factor back in to them right? – MSB Mar 24 '16 at 18:38
• $p(x)=1$ is not an irreducible quadratic, indeed not a quadratic at all. The two answers to your problem are $(x^3+x^2+1)(x^2+x+1)$ and $(x^3+x+1)(x^2+x+1)$. The first simplifies to $x^5+x+1$. You may be able to simplify the second without multiplying. – André Nicolas Mar 24 '16 at 18:46