# Show that polynomial is irreducible [duplicate]

I am trying to prove that the polynomial $P = X^5 + X^2 + 1 ∈ F_2 [X]$ is irreducible.

What I did:

I showed that $X^2+X+1∈F_2[X]$ is the only irreducible polynomial of degree 2. Is there a way to use this to prove that $P$ is irreducible without checking all the polynomial products giving polynomials of degree $5$?

Many thanks!

## marked as duplicate by Dietrich Burde abstract-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Apr 1 '18 at 16:07

• Well, sure. If you work $\pmod 2$ then either $P(x)$ has a root or it is divisible by the quadratic you mentioned. (or it's irreducible, of course). – lulu Apr 1 '18 at 15:38
• The answer of JavaMan for degree $5$ is very instructive; see the duplicate. – Dietrich Burde Apr 1 '18 at 16:08
Remark that if $P=Q_1Q_2$ the degree of $Q_1$ and $Q_2$ are different of $1$, then you can suppose that $Q_1=X^2+X+1$, but the roots of $X^2+X+1$ are third root of the unity, but a third root of the unity is not a root of $P$.
• The OP asks for the polynomial over $\Bbb F_2$ – Crostul Apr 1 '18 at 15:44
• @Crostul The trick with third roots of unity still works. If $a^3=1$, then $a^5+a^2=2a^2=0$, so $a$ is not a zero of $P$. I refrain from upvoting this (correct) solution, because the question has been handled many times already. – Jyrki Lahtonen Apr 1 '18 at 15:47