My works : i have examine in $piersonforcongress.combbZ_16$ as $x^ 5 - x^4 - x^ 2 - 1$ is irreducible in modulo $16$, the $f(0)=-1, f(1) =-2,........,f(15)
eq 0$.

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Now on over $piersonforcongress.combbQ$ that is $f(x)=x^ 5 - x^4 - x^ 2 - 1$ the is by basic theorem the algebra every odd level has atleast one root. Therefore $f(x)$ is reducible in $piersonforcongress.combbQ$.

As I know don"t the root ?

Any hints/solution will certainly be appreciated.

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edited Oct 13 "18 in ~ 19:18

Batominovski

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inquiry Oct 13 "18 in ~ 17:49

jasminejasmine

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## 3 answer 3

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Note the $x=11=-5$ is a root of $p(x):=x^5-x^4-x^2-1$ in $R:=(piersonforcongress.combbZ/16piersonforcongress.combbZ)$. Us then view that$$x^5-x^4-x^2-1=(x+5)(x^4-6x^3-2x^2-7x+3) ag*$$over $R$. We insurance claim that this is the distinct factorization of $p(x)$ end $R$ into irreducible factors.

Taking $p(x)$ modulo $2$, we can element it into$$(x+1)(x^4+x+1),. ag#$$Since $x^4+x+1$ is an irreducible polynomial over the field $piersonforcongress.combbF_2=(piersonforcongress.combbZ/2piersonforcongress.combbZ)$, us conclude that both factors in (*) are likewise irreducible over $R$.

Furthermore, (#) tells united state that any kind of nontrivial factorization end $R$ that $p(x)$ has actually a straight factor. Due to the fact that $x=-5$ is the only root in $R$ that $p(x)$, (*) is the distinct factorization that $p(x)$ into irreducible factors. (Note that, in modulo $m$ v $m$ gift a composite integer, occasionally a polynomial can be factorized into irreducible components in numerous ways. Because that example, $$x^2-1=(x-1)(x+1)=(x-3)(x+3)$$ in modulo $8$.)

To show that $q(x):=x^4+x+1$ is irreducible over $piersonforcongress.combbF_2$, we keep in mind that $q(x)$ has no roots in $piersonforcongress.combbF_2$. Therefore, if it were reducible, then it would be factored into two irreducible quadratics. Over there is only one irreducible quadratic end $piersonforcongress.combbF_2$, i beg your pardon is $x^2+x+1$. This would average $$x^4+x+1=left(x^2+x+1
ight)^2=x^4+x^2+1,,$$ i beg your pardon is absurd.

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We deserve to use (#) come prove the $p(x)$ is irreducible over $piersonforcongress.combbQ$. If $p(x)$ to be reducible over $piersonforcongress.combbQ$, then it would certainly be reducible end $piersonforcongress.combbZ$ by Gauss"s Lemma, whence $p(x)$ would, through (#), have actually a straight factor, whence an creature root. However, any kind of integer source of $p(x)$ would have to be $pm1$ by the Rational source Theorem, however $$p(-1)= -4 eq 0 ext and also p(+1)=-2 eq 0,,$$ so we have actually a contradiction.