Irreducible polynomials

May I ask what are Irreducible polynomials?
Examples with workings will be good.

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n_fortynineConnect With a Mentor Commented:
There is also a theorem that states that an irreducible polynimal p(x) in F[x] (where F is a field) will be reducible, i.e. have a root, in the extension field F[x]/p(x) (i.e. the field that contains all the remainders of a division of any polynomial in F[x] by p(x)). This theorem might not hold if F isn't a field.

A quick trick to recognize irreducibles of 2nd and 3rd degrees in F[x] is when they have no roots in F (F denotes a field).

For example: x^4 + x + 1 is irreducible in Z2[x], but has the root [x] in Z2[x]/(x^4 + x + 1) because [x]^4 + [x] + 1 = [x^4 + x + 1] = [0]

x^2 + x + 1 is also irreducible in Z2[x] but has the root [x^2 + x] in Z2[x]/(x^4 + x + 1) because [x^2 + x]^2 + [x^2 + x] + 1 = [x^4 + x + 1] = [0]

If you're unfamilar with rings, Z2 is the ring containing two elements [0] and [1], etc.

Hope this helps.
snoyes_jwConnect With a Mentor Commented:
Those polynomials that cannot be expressed as a product of non-trivial factors.  For example, x²-2 is irreducible over the set of rational numbers, because there are no rational numbers A and B such that x²-2 = (x+A)(x+B).

Straight from Google:
ozoConnect With a Mentor Commented:
A polynomial f(x) is irreducible over <R> if f(x) cannot be factoored as a product of polynomials in <R>[x] of degree less than the degree of f(x)
for example, the polynomial x²+1 is
irreducible in the Reals, because x²+1 has no Real root
reducible in the Complex field because x²+1 = (x-i)(x+i)
reducible in Z2 because x²+1 = (x+1)²
reducuble in Z5 because x²+1 = (x + 3)(x+2)
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