# For Wednesday, October 12th

For Wed:

2. For those of you who are still waiting on a paper re-do for the first assessment, we will do it in combination with the second assessment, so keep waiting a couple more days.
3. Let us consider polynomials in $x$ with coefficients in $\mathbb{Z}/3\mathbb{Z}$.   For example, $x + 2x \equiv 0$ and $2x^2 + 2x^2 \equiv x^2$.
1. Add $2x+2$ to $x^2 + 2x+1$.  The result should have coefficients chosen from $0, 1, 2$.
2. Multiply $x+2$ and $x+1$.  The result should have coefficients chosen from $0, 1, 2$.
4. Recall that we defined $\mathbb{Z}/n\mathbb{Z}$ as the set of integers with the additional “rule” that $n\equiv 0$ so we can remove multiples of $n$.  In the same way, consider the set of polynomials in the variable $x$ with coefficients in $\mathbb{Z}/3\mathbb{Z}$ (as in the last question), but “modulo” $x^2+1$.  In other words, whenever we see an $x^2+1$, we can remove it.  (Hint:  this also means whenever we see an $x^2$, we can replace it with $-1 \equiv 2$).  For example, $(x+1)(x+1) \equiv x^2 + 2x + 1 \equiv 2+ 2x + 1 \equiv 2x$.
1. Multiply out and simplify $(x+1)x$.
2. Multiply out and simplify $(x+1)(x+2)$.