Question

Find all solutions of the equation \( x^3-2x^2-3x=0\) in \(Z_{12}\)
Please, help

Answer 1

I got 0,3,4,8,11
but the answer from the back of the book is 0,3,5,8,9,11
I don't know how to get it

Answer 2

@UnkleRhaukus

Answer 3

the first step i would do is to factor out an x

Answer 4

yes, I did

Answer 5

x(x-3)(x+1)=0
so, 0, 3, and 11 are very first solution

Answer 6

why do you have to get the way you did it

Answer 7

then (3x4) =0 so, 4 is another one, right?

Answer 8

@Invert because I don't know other way, Please, show me your way

Answer 9

and then (3 x 8) =0 , too, --> 8 is another one,

Answer 10

What is \(Z_{12}\) ?

Answer 11

ring 12

Answer 12

{0,1,2,3,4,5,6,7,8,9,10,11} :)

Answer 13

ok, i have not studied rings, i dont think i can help you on this question

Answer 14

:( Thanks anyway XD

Answer 15

@abb0t

Answer 16

Idk. Sorry, bud.

Answer 17

It's ok, :(

Answer 18

Hate to admit it, but I don't have the slightest grasp of ring theory. I can tell you, however, that Winn66 is correct in his factoring of the given expression: x(x+1)(x-3). But the zeros of that are {0, 3, -1} (not 11).

Answer 19

-1 =11 in Z12, am I wrong?

Answer 20

I see that you recognize that that -1 needs to be converted into an alternative form for z12, and that you evidently know how to do that using the mod operator. Beyond that, Winner66, there's very little more that I could tell you. Very sorry.

Answer 21

I am sorry for not being here to get help. I have to go now. However, I really want to know how to solve the problem. Please, leave your instruction here. Much appreciate. Thanks in advance

Answer 22

\(\large x(x-3)(x+1) \equiv 0 \mod 12 \)

I would test one by one from x = 0 to 11
\(0(0-3)(0+1) \equiv 0\) ***
\(1(1-3)(1+1) \equiv -4\)
\(2(2-3)(2+1) \equiv 6\)
\(3(3-3)(3+1) \equiv 0\) ***
\(4(4-3)(4+1) \equiv 20 \equiv 8\)
\(5(5-3)(5+1) \equiv 5*12 \equiv 0\) ***
\(6(6-3)(6+1) \equiv 6*21 \equiv 6*3 \equiv 6\)
\(7(7-3)(7+1) \equiv 7*32 \equiv 7*8 \equiv -3\)
\(8(8-3)(8+1) \equiv 40*9 \equiv 0\) ***
\(9(9-3)(9+1) \equiv -3 * 6 * -2 \equiv 36 \equiv 0\) ***
\(10(10-3)(10+1) \equiv -2 * 7 * -1 \equiv 14 \equiv -2\)
\(11(11-3)(11+1) \equiv -1 * 8 * 12 \equiv 0\) ***

Answer 23

Thanks a ton. At the end, I know how to find them out mathematically. That's all I want to know. I <3 U. :)

Answer 24

u familiar wid solving quadratic congruences ?

Answer 25

nope xd

Answer 26

Sorry, I haven't studied rings yet. @loser66