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My understanding is to find out if a polynomial is reducible, we can use substitution. For example, to see if $x^2+1$ is reducible over $F_2$, we can substitute $0$ and $1$ for $x$, and see that neither of them result in $0$ mod $2$.

But we know $x$ is irreducible because it's of degree $1$ so you can't reduce it any further. Yet when you substitute $x$ for $0$, you get $0$. Does substituting $1$ have to yield $0$ as well?

So generally, when I say $x$ "is a root" in a field, given some polynomial over $x$, do I want to check that ALL substitutions for $x$ with the elements in the field give me $0$? or if there exists such an element such that the polynomial gives $0$?

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    That's only true for polynomials of degree $2,3$. For example, $x^4+4$ is reducible over $\mathbb Q$ by writing $x^4+4=(x^2-2x+2)(x^2+2x+2)$, but $x^4+4$ doesn't have any rational roots. – 2017-02-13
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    So does "irreducible over field $F$" mean that polynomial $\mathrm{deg\,}=k$ cannot be written as the product of polynomials $\mathrm{deg\,}$F$? – 2017-02-13

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A polynomial $f$ is irreducible over a field $\mathbb F$ if and only if it is an irreducible element of the ring $\Bbb F[x]$, id est, if and only if there are no $g,h\in\Bbb F[x]\setminus\{0\}$ such that $\deg g<\deg f$, $\deg h<\deg f$ and $gh=f$.

For instance, $x^2+1$ is not irreducible over $\Bbb F_2$ because $(x+1)\cdot(x+1)=x^2+1$ in $\Bbb F_2[x]$, while $x$ (and any polynomial of degree $1$) is irreducible over $\Bbb F_2$ because the product of two polynomials of degree strictly smaller than $1$ has degree $0$.

A polynomial has a root in $\Bbb F$ if and only if there is $\xi\in\Bbb F$ such that $f(\xi)=0$.

Due to the fact that, if $\Bbb F$ is a field, $x-\xi$ divides $f$ in $\Bbb F[x]$ if and only if $f(\xi)=0$ (sometimes it is referred to as Ruffini's theorem, I think), the following holds:

Let $\Bbb F$ be a field and let $f\in\Bbb F[x]$ with $1<\deg f\le 3$. Then, $f$ is irreducible over $\Bbb F$ if and only if it does not have roots in $\Bbb F$.

This "only if" part of the above does not extend to $\deg f>3$. For instance, in $\Bbb R[x]$, $$x^4+1=(x^2-\sqrt2x+1)(x^2+\sqrt2x+1)$$ but $\xi^4+1\ne 0$ for all $\xi\in\Bbb R$.

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    what is id est? – 2017-02-13
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    @Lana A Latin expression for "which means" (literally, "that is"). You may more commonly find it in English in the shorter version "i.e." – 2017-02-13
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    I'm not sure if that makes sense. $x^2+x+1$ is irreducible in F2. Neither substituting for 0 and 1 gives 0. – 2017-02-13
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    @Lana Of course it does not: I had written the opposite of what I meant. – 2017-02-13
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    Ok, so $x^2 + x + 1$ does not give 0. $x$ DOES give 0. So it has a root, but it's irreducable....? I'm really confused now – 2017-02-13
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    No. $x^2+x+1$ does not have roots in $\Bbb F_2$ because $0^2+0+1=1=1^2+1+1$. And, in fact, it is irreducible over $\Bbb F_2$. – 2017-02-13
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    That's what I said. The polynomial X has roots. x=0. And its irreducible – 2017-02-13
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    Yes, but its degree is neither $2$ nor $3$. – 2017-02-13
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    ohhhh ok, so this is literally just deg. 2 and 3. Interesting. – 2017-02-13