A polynomial can not be the product of two polynomials of degree 2 and 3 I want to show that 
the polynomial $x^5+x^2-1$ in $\Bbb{Z}/2\Bbb{Z}[x]$ can not  be written can a product of a polynomial of degree $2$ and an other one of degree $3$.
For that I wrote $x^5-x^2+1=(x^2+bx+c)(x^3+dx^2+ex+f)=x^5+(b+d)x^4+(c+e+bd)x^3+(f+cd+be)x^2+(bf+ce)x+cf$
where $a,b,c,d,e,f\in  \Bbb{Z}/2\Bbb{Z}$
then by identification we have
$b+d=0 \\
c+e+bd=0\\
f+cd+be=-1\\
bf+ce=0\\
cf=1$
but I didn't find a contradiction, can you please help me? Thanks
 A: First off, the line $cf=1$ tells us that $c = f = 1$. Now insert that into the rest of the equaitons (along with substituting $-1 = 1$ because $-$ is meaningless over $\Bbb Z_2$):
$$b+d=0 \\
1+e+bd=0\\
1+d+be=1\\
b+e=0$$
From the first and last equations, we get that $d = b = e$, so we insert that into the final two equations, and we get
$$
1+b+b^2=0\\
1+b+b^2=1$$
and there you have your contradiction.
A: Let us show that the polynomial $x^5-x^2+1$ is irreducible in $\Bbb Z[x]$.
We khnow that if a polynomial in reducible in $\Bbb Z[x]$ then it is reducible in $\Bbb{Z}/p\Bbb{Z}[x]$ for any prime number $p$.
Therefore if we show for example that $x^5+x^2+1$ is irreducible in $\Bbb{Z}/2\Bbb{Z}[x]$  then $x^5-x^2+1$ is irreducible in $\Bbb Z[x]$.
So let us show that $x^5+x^2+1$ is irreducible in $\Bbb{Z}/2\Bbb{Z}[x]$ :
Since $0^5+0^2+1=1\not=0$ and $1^5+1^2+1=1\not=0$ then we conclude that $x^5+x^2+1$ does not have any linear factor in $\Bbb{Z}/2\Bbb{Z}[x]$.
Now we need to show that $x^5+x^2+1$ can not be written as a product of a polynomials of degree 2 and 3.
For that let us suppose that there exist $a,b,c,d,e,f\in  \Bbb{Z}/2\Bbb{Z}$
such that  $\begin{align*}
x^5+x^2+1&=(x^2+bx+c)(x^3+dx^2+ex+f)\\&=x^5+(b+d)x^4+(c+e+bd)x^3+(f+cd+be)x^2+(bf+ce)x+cf\end{align*}$
then by identification we have
$$\left\{
    \begin{array}{ll}
b+d=0 \\
c+e+bd=0\\
f+cd+be=1\\
bf+ce=0\\
cf=1 \end{array}
\right.$$
First off, the line $cf=1$ tells us that $c = f = 1$. Now insert that into the rest of the equations:
$$\left\{
    \begin{array}{ll}b+d=0 \\
1+e+bd=0\\
1+d+be=1\\
b+e=0\end{array}
\right.$$
From the first and last equations, we get that $d = b = e$, so we insert that into the final two equations, and we get
$$\left\{
    \begin{array}{ll}
1+b+b^2=0\\
1+b+b^2=1\end{array}
\right.$$
and there we have a contradiction.
