Encode the message $[1,1,0,1,1,0,1]$ in BCH code based on the field $\mathbb F = \frac{\mathbb Z_{2}[x]}{x^4+x+1}$ So here's what I understand so far:
$\mathbb F = \frac{\mathbb Z_{2}[x]}{x^4+x+1} = GF(16)$
The code is written as $[x^{14},x^{13},x^{12},x^{11},x^{10},x^{9},x^{8}$ $|$ $x^{7},x^{6},x^{5},x^{4},x^{3},x^{2},x^{1},x^{0}]$
So we write the message $[1,1,0,1,1,0,1] = x^{14}+x^{13}+x^{11}+x^{10}+x^{8}$ and we get the first part of the code:
$[1,1,0,1,1,0,1$ $|$ $x^{7},x^{6},x^{5},x^{4},x^{3},x^{2},x^{1},x^{0}]$
We divide $x^{14}+x^{13}+x^{11}+x^{10}+x^{8}$ by $(x^8+x^7+x^6+x^4+1)$ and we get $(x^6+x^4+x^2+x)$ with remainder $(x^7+x^5+x^4+x^2+x)$
Taking the coefficients of the remainder we get the encoded BCH code
$[1,1,0,1,1,0,1$ $|$ $1,0,1,1,0,1,1,0]$ 
My question is: Why do we divide by $(x^8+x^7+x^6+x^4+1)$? Where did that come from?:

Let $\alpha$ be a root of $m_{\alpha}(x)=x^4+x+1$ in the field $GF(16)$. As this polynomial is primitive, $\alpha$ is of order $15$. Therefore $\alpha^3$ is  of order five. Thus the minimal polynomial of $\alpha^3$ is $m_{\alpha^3}(x)=x^4+x^3+x^2+x+1$. The generator polynomial of a double-error-correcting BCH-code is
$$
m_{\alpha}(x)m_{\alpha^3}(x)=(x^4+x+1)(x^4+x^3+x^2+x+1)=x^8+x^7+x^6+x^4+1.
$$
The rest of the calculation is basic use of the generator polynomial that has undoubtedly been explained in you textbook/lecture notes.
Related topics have been discussed in another question, see Dilip Sarwate's answer and the Wikipedia article linked to in that question. It is about Reed-Solomon codes, but the algebra is more or less the same as in the case of BCH-codes.
