Let $c(x) = \sum_{i=0}^{N-1} c_i x^i$ denote the polynomial corresponding
to the codeword $\mathbf c = [c_0, c_1, \ldots, c_{N-1}]$ of a cyclic binary Hamming
code with generator polynomial $g(x)$. Note that $N = 2^m-1$ for some
positive integer $m$ and that $g(x)$ is a primitive
polynomial of degree $m$. Let $\alpha \in \mathbb F_{2^m}$ be a root of
$g(x)$ and note that $\alpha$ has order $2^m-1$.
The parity check matrix $H$ of this cyclic binary Hamming code can be viewed as
$[1, \alpha, \cdots, \alpha^{N-1}]$. If you are unfamiliar with this notion, then
express each $\alpha^i$ as a column vector to get the
usual binary representation. But, with this formulation,
$H\mathbf c^T = \mathbf0$ is equivalent to $c(\alpha) = 0$ which is
exactly what the polynomial formulation says in other terms: each
codeword polynomial is a multiple of $g(x)$ which happens to have
$\alpha$ as a root, and so $c(\alpha) = 0$.
Let $\mathbf c$ be transmitted and $\mathbf r = \mathbf c + \mathbf e$ denote
the received word. In terms of polynomials, $r(x) = \sum_i r_ix^i$
is the received word polynomial where $r_i = c_i + e_i$ with
$e(x)$ denoting the error word polynomial corresponding to the error
word $\mathbf e$. Note that $r(x) = c(x) + e(x)$. The receiver computes
$$H\mathbf r^T = H(\mathbf c + \mathbf e)^T =
H\mathbf c^T + H\mathbf e^T = \mathbf 0 + H\mathbf e^T = H\mathbf e^T.$$
In terms of polynomials, computing $H\mathbf r^T$ is the same as
evaluating $r(x)$ at $\alpha$, andso the above equation can be
expressed as
$$r(\alpha) = c(\alpha) + e(\alpha) = 0 + e(\alpha) = e(\alpha).$$
If a single error has occurred, then $\mathbf e$ has Hamming weight $1$,
and $e(x)$ is a monomial $x^i$ for some $i$, $0 \leq i \leq N-1$.
Thus, the bit in the $i$-th position has been received incorrectly.
In terms of polynomials, the syndrome is
$e(\alpha) = x^i\bigr |_{x=\alpha^i} = \alpha^i$. If two errors
have occurred, the syndrome is $\alpha^i + \alpha^{i^\prime} = \alpha^j$.
The decoder for the cyclic Hamming code cannot tell whether one
or two errors have occurred: if the symdrome equals $\alpha^\ell$,
it simply inverts $r_\ell$ to correct the error.
The extended Hamming code has an overall parity bit $c_N$
whose value is chosen to ensure that the extended codeword
$\hat{\mathbf c} = (\mathbf c, c_N)$ has even Hamming weight.
Suppose that an extended Hamming code is being used and
it is known that if there are two channel errors,
then they are adjacent errors not involving the overall
parity bit.
Let $\hat{\mathbf r} = (\mathbf r, r_N)$ be the received word for
the extended Hamming code. If $\hat{\mathbf r}$ has odd Hamming
weight, then the decoder assumes that a single error has
occurred, and if $r(\alpha) = \alpha^\ell$, the decoder inverts
$r_\ell$.
But if $\hat{\mathbf r}$ has even Hamming weight and
$r(\alpha) = \alpha^\ell \neq 0$, then the decoder assumes that
$e(x) = x^i + x^{i+1}$ (double adjacent error) and so
$r(\alpha) = e(\alpha) = \alpha^i + \alpha^{i+1} = \alpha^\ell.$
Since $\alpha^i = \alpha^\ell(1+\alpha)^{-1}$, the
decoder can find the value of $i$ and then invert $r_i$ and $r_{i+1}$.
In this way, single errors and double adjacent errors not involving
the overall parity bit can be corrected with an extended Hamming code.
Best Answer
The generator polynomial of a cyclic binary $[2^n-1, 2^n-1-n]$ Hamming code is always a primitive polynomial of degree $n$. So, use one of the two primitive polynomials of degree $4$ that you have exhibited as factors of $x^{15}-1$. (Hint: you may need to first figure out which two of the three polynomials of degree $4$ are primitive and which one is nonprimitive).
Thus there are two different cyclic binary $[15,11]$ Hamming codes depending on which primitive polynomial you choose as the generator polynomial. They are related in the sense that if $(C_0, C_1, \ldots, C_{14})$ is a codeword in one code, then $(C_{14}, C_{13}, \ldots, C_1,C_0)$ is a codeword in the other code.