Question #200238

Find the solution of the equation x3-2x2-3x in z12

1
Expert's answer
2021-05-31T17:41:02-0400

First let us factorize this polynomial :

x32x23x=x(x3)(x+1)x^3-2x^2-3x=x\cdot(x-3)\cdot(x+1)

The obvious solutions are 0;3;110; 3; 11 (as 11=111=-1 in Z12\mathbb{Z}_{12} ). However, as Z12\mathbb{Z}_{12} contains the zero divisors, there could possibly be other solutions.

12=4312=4\cdot 3 and thus if xx is a solution of this equation, we should have 3x3|x or 3(x+1)3|(x+1) (as the condition 3(x3)3|(x-3) is equivalent to 3x3|x). Therefore, the other possible solutions may be 6;9;2;5;86;9;2;5;8

Now we can also see, that if xx is even, then x3,x+1x-3, x+1 are odd and thus if xx is a solution and it is even, it should be divisible by 4. However, if xx is odd, then both x3,x+1x-3, x+1 are even, so their product is divisible by 4. These conditions yield that the solution set is

S:={0;3;5;8;9;11}S:= \{0;3;5;8;9;11 \}


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