My works : ns have check in $mmsanotherstage2019.combbZ_16$ as $x^ 5 - x^4 - x^ 2 - 1$ is irreducible in modulo $16$, the $f(0)=-1, f(1) =-2,........,f(15) eq 0$.

You are watching: Factor (x^2+1)

Now on over $mmsanotherstage2019.combbQ$ the is $f(x)=x^ 5 - x^4 - x^ 2 - 1$ that is by fundamental theorem of algebra every odd degree has atleast one root. So $f(x)$ is reducible in $mmsanotherstage2019.combbQ$.

As I understand don"t the root ?

Any hints/solution will be appreciated.

abstract-algebra polynomials factoring irreducible-polynomials finite-rings
edited Oct 13 "18 at 19:18

47.8k44 yellow badges4848 silver- badges120120 bronze badges
request Oct 13 "18 in ~ 17:49

| present 2 more comments

3 answers 3

energetic oldest Votes
Note that $x=11=-5$ is a root of $p(x):=x^5-x^4-x^2-1$ in $R:=(mmsanotherstage2019.combbZ/16mmsanotherstage2019.combbZ)$. 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 case that this is the unique factorization of $p(x)$ over $R$ right into irreducible factors.

Taking $p(x)$ modulo $2$, us can variable it into$$(x+1)(x^4+x+1),. ag#$$Since $x^4+x+1$ is one irreducible polynomial over the ar $mmsanotherstage2019.combbF_2=(mmsanotherstage2019.combbZ/2mmsanotherstage2019.combbZ)$, us conclude the both determinants in (*) are additionally irreducible end $R$.

Furthermore, (#) tells united state that any kind of nontrivial factorization over $R$ that $p(x)$ has a linear factor. Because $x=-5$ is the just root in $R$ of $p(x)$, (*) is the distinctive factorization that $p(x)$ into irreducible factors. (Note that, in modulo $m$ v $m$ being a composite integer, sometimes a polynomial deserve to be factorized right into irreducible components in many ways. For 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 end $mmsanotherstage2019.combbF_2$, we keep in mind that $q(x)$ has actually no roots in $mmsanotherstage2019.combbF_2$. Therefore, if it to be reducible, then it would be factored into two irreducible quadratics. There is only one irreducible quadratic end $mmsanotherstage2019.combbF_2$, i m sorry 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.

See more: Macon To Atlanta Distance From Atlanta Georgia To Macon Georgia

We can use (#) come prove the $p(x)$ is irreducible end $mmsanotherstage2019.combbQ$. If $p(x)$ to be reducible end $mmsanotherstage2019.combbQ$, climate it would certainly be reducible over $mmsanotherstage2019.combbZ$ by Gauss"s Lemma, whence $p(x)$ would, by (#), have a direct factor, whence an essence root. However, any kind of integer root of $p(x)$ would have to be $pm1$ by the Rational source Theorem, but $$p(-1)= -4 eq 0 ext and also p(+1)=-2 eq 0,,$$ therefore we have actually a contradiction.