Solve and factorize in Mod.6: x^5 - 3x^4 - 2x^2 + x?

Answer 1

We can take out x as a factor to get x(x^4 -3x^3 -2x + 1).
We still have to prove that the last factor, p, is irreducible
mod 6.
First, let's check for linear factors, using the
factor theorem:
x = 2 or 4 gives an odd number, which cannot be a
root mod 6.
x = 3 gives a number congruent to 1(mod 3), so cannot
be a root mod 6.
Finally, x = 1 or -1 do not give a zero mod 6.
So there are no linear factors mod 6.
Thus the only possibility is 2 quadratic factors.
So suppose p = ( x^2 +ax + b)(x^2 + cx + d) (mod 6)
(Yes, both leading factors can also be 5, but the
analysis is the same.)
This says that bd = 1(mod 6). The only invertible
elements mod 6 are 1 and 5.
So either b = d = 1 or b = d = 5 (mod 6).
Suppose first b = d = 1.
Equating the x^3 terms gives a + c =3(mod 6)
Equating the x terms gives a + c = 4(mod 6),
a contradiction.So this case is impossible.
If b = d = 5(mod 6), we get exactly the same
contradiction. So p is irreducible (mod 6).
Hope I've answered the question you asked!

Answer 2

x can be taken as a factor
x(x^4-3x^3-2x+1)