The word $r(x)=1+x^3+x^9$ is received while using a binary narrow sense BCH code of length 15 and designed distance $\delta=5$, find the correct word being sent.
I am following the method on Wiki to do this problem, but I don't know how to find $e$, so far I have the following work:
The BCH code is $\mathcal{B}_{(q=2)}(15,5,\omega,1)$ and its generator polynomial is $g(x)=(x^4+x+1)(x^4+x^3+x^2+x+1)=x^8+x^7+x^6+x^4+1$. Roots of $g$ are $\omega,\omega^2,\omega^3,\omega^6,\omega^8,\omega^9,\omega^{12}$. Let $\omega$ be the root of $x^4+x+1$, we have the following table, \begin{array}{c|c|c|c} i & \omega^i & i & \omega^i\\ \hline 0 & 1 & 8 & \omega^2+1\\ 1 & \omega & 9 & \omega^3+\omega\\ 2 & \omega^2 & 10 & \omega^2+\omega+1\\ 3 & \omega^3 & 11 & \omega^3+\omega^2+\omega\\ 4 & \omega+1 & 12 & \omega^3+\omega^2+\omega+1\\ 5 & \omega^2+\omega & 13 & \omega^3+\omega^2+1\\ 6 & \omega^3+\omega^2 & 14 & \omega^3+1\\ 7 & \omega^3+\omega+1 & & \end{array}
As $\delta=5$ is designed distance which is minimum distance, so $t=2$. Since $r(x)=1+x^3+x^9$ is the received massage, the syndromes are \begin{align*} S_1 & = 1+\omega^3+\omega^9=1+\omega^3+(\omega^3+\omega)=\omega+1=\omega^4=r(\omega)\\ S_2 & = 1+\omega^6+\omega^{18}=1+(\omega^3+\omega^2)+\omega^2\omega^9=1+(\omega^3+\omega^2)+\omega^2(\omega^3+\omega)=\omega^2+1=\omega^8=r(\omega^2)\\ S_3 & = 1+\omega^9+\omega^{27}=1+(\omega^3+\omega)+\omega^3\omega^9=1+(\omega^3+\omega)+\omega^3(\omega^3+\omega)=\omega^2=r(\omega^3)\\ S_4 & = 1+\omega^{12}+\omega^{36}=1+\omega^3\omega^4+\omega^4\omega^9=1+\omega^3(\omega+1)+(\omega+1)(\omega^3+\omega)=\omega^2+\omega+1=\omega^{10}=r(\omega^4) \end{align*}
Can someone give me a hint or suggestion to find $e$? Thanks