# reduces the coefficients of a modulo 3 on NTRU

i'm still don't understand about 'reduces the coefficients of a modulo 3' on NTRU tutorial

$a = f*e = 3 - 7X - 10X^2 - 11X^3 + 10X^4 + 7X^5 + 6X^6 + 7X^7 + 5X^8 - 3X^9 -7X^{10} \pmod{32}.$ Note that when Bob reduces the coefficients of $f*e \bmod{32}$, he chooses values lying between -15 and 16, not between 0 and 31. It is very important that he choose the coefficients in this way.

Next Bob reduces the coefficients of a modulo 3 to get $b = a = - X - X^2 + X^3 + X^4 + X^5 + X^7 - X^8 - X^{10} \pmod{3}$.

how can you get polynomial $a \bmod{3}$ is : $- X - X^2 + X^3 + X^4 + X^5 + X^7 - X^8 - X^{10}$? what i know on 'polynomial $a \bmod{3}$' means every coefficient mod 3, so coefficient on $-X^8$ should be $2$ ($2X^8$) not $-1$ ($-X^8$), because $5 \bmod 3 = 2$?

$-7X$ becomes $-X$ because $-7 \equiv -1 \mod 3$. The starting constant 3, becomes 0, and disappears, $-10X^2$ becomes $-X^2$ (adding 9), $-11X^3$ becomes $X^3$, as $-11 \equiv 1 \mod 3$ (add 12 to the left) etc. etc.
Your final example $5X^8$ could become $2X^5$, but also (subtracting 3 again, to get a representative from $\{-1,0,1\}$) the actually used $-X^8 = (-1)X^8$.