Factor $x^ 5 - x^4 - x^ 2 - 1$ modulo $16$, and over $\mathbb{Q }$ From ARTIN algebra books chapter $12$ question $4.19$:
Factor $x^ 5 - x^4 - x^ 2 - 1$  modulo $16$, and over $\mathbb{Q }$ 
My works : I have check in $\mathbb{Z}_{16}$ as $x^ 5 - x^4 - x^ 2 - 1$ is  irreducible in modulo $16$, that $f(0)=-1, f(1) =-2,........,f(15)\neq 0$.
Now on over $\mathbb{Q}$  that 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 $\mathbb{Q}$.
As I know don't the root ?
Any  hints/solution will be appreciated.
 A: Note that $x=11=-5$ is a root of $p(x):=x^5-x^4-x^2-1$ in $R:=(\mathbb{Z}/16\mathbb{Z})$. We then see that
$$x^5-x^4-x^2-1=(x+5)(x^4-6x^3-2x^2-7x+3)\tag{*}$$
over $R$.  We claim that this is the unique factorization of $p(x)$ over $R$ into irreducible factors.
Taking $p(x)$ modulo $2$, we can factor it into
$$(x+1)(x^4+x+1)\,.\tag{#}$$
Since $x^4+x+1$ is an irreducible polynomial over the field $\mathbb{F}_2=(\mathbb{Z}/2\mathbb{Z})$, we conclude that both factors in (*) are also irreducible over $R$.  
Furthermore, (#) tells us that any nontrivial factorization over $R$ of $p(x)$ has a linear factor.  Since $x=-5$ is the only root in $R$ of $p(x)$, (*) is the unique factorization of $p(x)$ into irreducible factors.  (Note that, in modulo $m$ with $m$ being a composite integer, sometimes a polynomial can be factorized into irreducible factors 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 over $\mathbb{F}_2$, we note that $q(x)$ has no roots in $\mathbb{F}_2$.  Therefore, if it were reducible, then it would be factored into two irreducible quadratics.  There is only one irreducible quadratic over $\mathbb{F}_2$, which is $x^2+x+1$.  This would mean $$x^4+x+1=\left(x^2+x+1\right)^2=x^4+x^2+1\,,$$ which is absurd.


We can use (#) to prove that $p(x)$ is irreducible over $\mathbb{Q}$.  If $p(x)$ were reducible over $\mathbb{Q}$, then it would be reducible over $\mathbb{Z}$ by Gauss's Lemma, whence $p(x)$ would, by (#), have a linear factor, whence an integer root.  However, any integer root of $p(x)$ would have to be $\pm1$ by the Rational Root Theorem, but $$p(-1)= -4\neq 0\text{ and }p(+1)=-2\neq 0\,,$$ so we have a contradiction.
A: $f=x^5-x^4-x^2-1$ is irreducible over $\mathbb{F}_3$, and therefore by the reduction theorem irreducible over $\mathbb{Q}$. To see this, write 
$$
f=(x^3+ax^2+bx+c)(x^2+dx+e)
$$
and compare coefficients. The equations do not have a solution in $\mathbb{F}_3$. 
Since also $f$ has no root there, we are done.
A: Let $f=x^5-x^4-x^2-1$.

As noted in the comments, $f(11)\equiv 0\;(\text{mod}\;16)$, so $f$ is reducible, mod $16$.

Suppose $f$ is reducible in $\mathbb{Q}[x]$.

Then $f$ must also be reducible in $\mathbb{Z}[x]$, hence, since $f$ is monic, $f=gh$, for some monic, non-constant polynomials $g,h\in\mathbb{Z}[x]$.

Without loss of generality, assume $\deg(g)\le \deg(h)$.

By the rational root test, $f$ doesn't have a rational root, hence $\deg(g) > 1$.

Then we must have $\deg(g)=2$, so $g=x^2+ax+b$, for some $a,b\in\mathbb{Z}$.

Since the constant term of $f$ is $-1$, it follows that $|b|=1$.

Consider two cases . . .

Case $(1)$:$\;b=1$.

Then $g=x^2+ax+1$.

By polynomial long division, dividing $f$ by $g$, the remainder has constant term
$$a^3+a^2-2a-1$$
contradiction, since by the rational root test, the equation
$$a^3+a^2-2a-1=0$$
has no rational root.

Case $(2)$:$\;b=-1$.

Then $g=x^2+ax-1$.

By polynomial long division, dividing $f$ by $g$, the remainder has constant term
$$-a^3-a^2-2a-3$$
contradiction, since by the rational root test, the equation
$$-a^3-a^2-2a-3=0$$
has no rational root.

It follows that $f$ is irreducible in $\mathbb{Q}[x]$.
