In galois field- how does modular arithmetic

Question

The following is given in the text that I have: In $GF(2^8)$ [Galois Field] let: $$h(x)=x^8+x^4+x^3+x+1$$ $$x^8 \bmod h(x)= [h(x)-x^8]$$ I basically don't understand the second step I think $h(x) mod x^8=[h(x)-x^8]$, is the text mistaken or am I ? If I am please explain this to me.

Answer 1

By definition, $x^8\equiv x^8-h(x) \bmod h(x)$, namely $h(x)\mid (x^8+h(x)-x^8)$. On the other hand, since $2=0$ we have $h(x)-x^8=x^8-h(x)$.

Answer 2

The leading terms of $x^8$ and $h(x)$ are the same so that the quotient is $1$, and both

$$x^8=h(x)+(x^8\bmod h(x))$$ and $$h(x)=x^8+(h(x)\bmod x^8)$$ hold.