1. ## Find zeros

Show that $x^2 + 3x + 2$ has four zeros in $Z_{6}$.

$x^2 + 3x + 2 = (x+2)(x+1)$, by the factor theorem, -1 and -2 are zeros of the poly.

However, -2 = 4 and -1 = 5 in $Z_{6}$. But what are the other two?

Thanks.

Show that $x^2 + 3x + 2$ has four zeros in $Z_{6}$.

$x^2 + 3x + 2 = (x+2)(x+1)$, by the factor theorem, -1 and -2 are zeros of the poly.

However, -2 = 4 and -1 = 5 in $Z_{6}$. But what are the other two?

Thanks.

since $Z_6$ is a finite set, if you use tables, you can find it easily..

----0---1---2---3---4---5
f(x) 2---0---0---2---0---0

3. I just want to mention the interesting fact that even though the polynomial is of degree 2 it has 4 zeros. That might seem impossible but there is a reason for it. That theorem (about polynomials having at most the number of zeros of its degree) is only for fields (or when the division algorithm works). But Z_6 is not a field so it is not supprising why it fails.

4. I see, so really, I just have to plug the numbers in.

The zeros are 1 , 2 , 4 , 5.

Thanks.