# Factorize Polynomials

Polynomials over the body $Z_2$ are viewed.

Determine for $p(z) = z^4 + z^3 + z$ and $q(z) = z^2 + 1$

1. $f(z) = p(z) + q(z)$
2. $g(z) = p(z) * q(z)$
3. $h \equiv p\ mod\ q$ with minimal degree(h)

And factorize the polynomials if possible

What i did was:

1. $f(z) = z^4 + z^3 + z^2 + z + 1$
2. $g(z) = z^6 +z^5 +z^4 + z^3 + z^3 + z$
3. $h \equiv (z^4 + z^3 + z):(z^2 + 1) = z^2 + z +$${-z^2}\over{z^2 + 1} h \equiv -z^2 My question is: How should i factorize the polynomials? I dont think i can factorize it more! And is 3 correct? Thanks - Modulo 2, q(z) = (z+1)^{2} – neelp Dec 29 '13 at 14:28 About question 2: What is z^3+z^3 \mod 2? – TonyK Jan 6 '14 at 9:00 (z^4 + z^3 + z) \equiv (z^2 + 1) *(z^2 + z +1)+1\pmod{2}, so h \equiv 1 \pmod{2} – miracle173 Jan 6 '14 at 23:47 ## 1 Answer Factorizing in \mathbb Z_2 is fun. Let's try f. It has a linear factor iff it has a root. It doesn't. So if it factors, it factors into two quadratic, say:$$ (ax^2+bx+c)(dx^2+ex+F)$$Now$ad=1$, so$a=d=1$.$ae+bd=b+e=1$, so$b\neq e$. Similarly,$c=F=1$. Next,$aF+be+cd=be=1$, so$b=e=1$. This is a contradiction. So$f\$ can't be factored.

There's probably a less stupid way to do this, but those ways are boring.

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Thanks! Im not sure if i get you 100%! Could you please give me a example with my numbers? Thanks – John Smith Dec 29 '13 at 14:36
This was with your numbers... – Gaffney Dec 29 '13 at 14:37