Your calculations are correct.
It is worth keeping in mind that the quotient is not of importance in the CRC calculations, it is the remainder that is needed. The careful calculations
that you have carried out and written up in detail are good for familiarizing
oneself with the algorithm. However, after some practice, you may find the
following shorthand much more handy
010000
1101
1101
111
I believe your doubt is not really related to CRC but with cyclic codes (on which CRC are based), more specifically with the construction of systematic cyclic codes. This is explained in any textboox. Here's a summary.
A binary $(n,k)$ cyclic code has $2^k$ codewords, they correspond to a binary polynomial $c(X)$ of degree less than $n$. A particular code is specified by a generator polynomial $g(X)$, which must be of degree $n-k$ and must be a factor of $X^n+1$.
The direct (non-systematic) coding is then $$c(X)=m(X)g(X) \tag{1}$$ where $m(X)$ represents the raw message (as a polynomial of degree less than $k$).
The decoding is done by dividing the received message $r(X)=c(X)+e(X)$ by the generator $g(X)$ and looking at the remainder $s(X)$ ("syndrome", degree less than $n-k$) :
$$r(X)=g(X)q(X)+s(X) \tag{2}$$
If there were no errors ($e(X)=0)$ then the $s(X)=0$
An slightly different procedure produces a systematic coding. Here, to code, we compute the remainder of $X^{n-k}m(X)$ divided by $g(X)$:
$$ X^{n-k}m(X)=q(X)g(X)+p(X) \tag{3}$$
or, equivalently
$$ X^{n-k}m(X)+p(X)=q(X)g(X) \tag{4}$$
Because the left side is a multiple of $g(X)$, that must be one of the codewords of our cyclic code (equivalent to the previous one; but, mind you, it's not the same: it has the same codewords, but mapped to different input messsages). This code has the nice property of being the concatenation of the original raw message $m(X)$ shifted $n-k$ positions, and the "parity polynomial" $p(x)$ (degree less than $n-k$).
So, at the end you trasmit $c(X)= X^{n-k}m(X)+p(X)$, i.e. the sum (which here amounts to a concatenation) of the raw message shifted $n-k$ positions and the remainder $p(X)$.
The decoding is the same as before: just compute the remainder of the division by $g(x)$.
CRCs are essentially systematic cyclic codes, in which the parity length $n-k$ is relatively small (typically 16 or 32) and $n$ is large (say $n=2^{15}-1=32767$); however, one does not usually code the full length block, but a smaller message of arbitrary length, to which the computed parity bits will be appended. As explained here, this is mathematically equivalent to assuming that the unused bits of the total block are zero.
Best Answer
the vonbrand answer is slightly off. This is the correct operations in GF(2) -
In GF(2) X^10+X^7+X^6+X^4 MOD (X^3+X^2+1) = X^2+1
Therefor your text book's answer is correct.