Factor 2x^3 + 4x + 1 in Z_5[x] (modulo 5)
I can't figure this out. Can someone show please? Thanks!
I'm sure there's a more general way to do this, but as we're talking about $\displaystyle \mathbb{Z}_5$ there are only 5 possibilities so it's not that hard to do a "chug and plug." If you want a bit more rigor, then note that $\displaystyle 1 \equiv -4$:
$\displaystyle 2x^3 + 4x + 1 \equiv 2x^3 + 4x - 4 \equiv 0$
Since 5 is a prime there is a unique inverse of 2 so we can multiply both sides by 3 and get
$\displaystyle x^3 + 2x - 2 \equiv 0$
The last two terms are even, so thus $\displaystyle x^3$ needs to be as well. Thus the only possible values for x are 0, 2, and 4.
-Dan
everything up to here is true. but "even" has no meaning in Z_{5}: 1 = 2+2+2. furthermore "0" in Z_{5} might mean some ODD multiple of 5 (as an ordinary integer).
however, it IS indeed true that: (2)(2^{3}) + (4)(2) + 1 = 0 (mod 5) and 2(4^{3}) + (4)(4) + 1 = 0 (mod 5).
i find that using a substitution like 1 = -4 (mod 5) can get confusing...it leads one to use properties of integers that DO NOT HOLD in Z_{5}.
rather, multiply the original equation by 3:
3(2x^{3} + 4x + 1) = x^{3} + 2x + 3 = 0 (in Z_{5}) <--this is a monic polynomial, so we're looking for monic factors.
we have two roots in Z_{5}, so it must factor completely over Z_{5}, so one of these (2 or 4) is a repeated root. we COULD find the repeated root by long division but meh...i'm lazy.
so let's take the derivative! we get:
3x^{2} + 2, and if we set this equal to 0 and multiply by 2, we get:
x^{2} + 4 = 0 <---this has the roots x = 1 and x = 4, the shared root with x^{3} + 2x + 3 is x = 4, so 4 is our repeated root.
thus:
x^{3} + 2x + 3 = (x + 1)^{2}(x + 3) (i am writing x + 1 instead of x - 4, and x + 3 instead of x - 2), so:
2x^{3} + 4x + 1 = 2(x + 1)^{2}(x + 3), as can be verified by multiplying the factors (you could also write this as 2(x - 4)^{2}(x - 2), if you like seeing the roots in your factors).