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I have the following polynomial

$$g(x)=x^6+x^5+x^4+x^2 \in ℤ \bmod 2$$

I'm having trouble finding information on how
to reduce this polynomial, Or let alone being able to tell if it is reducible in the first place.

If the polynomial has a root then it is reducible correct So
$$g(0)=0^6+0^5+0^4+0^2=0 \in ℤ \bmod 2$$

$$g(1)=1^6+1^5+1^4+1^2=0 \in ℤ \bmod 2$$

So it is reducible?

EDIT: Okay it's reducible I guess i'm just confused on how to actually reduce it now.

Michael Hardy
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Temirzhan
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  • $x^6+x^5=x^5(x+1)=0$ and $x^4+x^2=x^2(x^2+1)=x^2(x+1)^2=0$ and therefore you have the zero polynomial over $\Bbb{Z}_2.$. – Bumblebee Nov 07 '17 at 03:22
  • This is NOT the zero polynomial. It's true its only value as function on $\mathbb{Z}/2$ is 0, but that we knew already from the OP. – Torsten Schoeneberg Nov 07 '17 at 03:48

1 Answers1

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You can begin with factoring out $x^2$: $$x^6+x^5+x^4+x^2=x^2(x^4+x^3+x^2+1).$$ Now observe that $w=1$ is a root of the second factor, so it is divisible by $x-1=x+1$ (we're in characteristic $2$). Indeed, using synthetic division, we readily obtain $$x^6+x^5+x^4+x^2=x^2(x+1)(x^3+x+1),$$ and this last factor is irreducible since it has degree $3$ and no root in $\mathbf F_2$.

Bernard
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