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About LinkBacksFactor 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 aboutOriginally Posted by jzellt
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
:
Since 5 is a prime there is a unique inverse of 2 so we can multiply both sides by 3 and get
The last two terms are even, so thusneeds 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 Z5: 1 = 2+2+2. furthermore "0" in Z5 might mean some ODD multiple of 5 (as an ordinary integer).The last two terms are even, so thus
-Dan
however, it IS indeed true that: (2)(23) + (4)(2) + 1 = 0 (mod 5) and 2(43) + (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 Z5.
rather, multiply the original equation by 3:
3(2x3 + 4x + 1) = x3 + 2x + 3 = 0 (in Z5) <--this is a monic polynomial, so we're looking for monic factors.
we have two roots in Z5, so it must factor completely over Z5, 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:
3x2 + 2, and if we set this equal to 0 and multiply by 2, we get:
x2 + 4 = 0 <---this has the roots x = 1 and x = 4, the shared root with x3 + 2x + 3 is x = 4, so 4 is our repeated root.
thus:
x3 + 2x + 3 = (x + 1)2(x + 3) (i am writing x + 1 instead of x - 4, and x + 3 instead of x - 2), so:
2x3 + 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).
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