Polynomial representation of binary It is well known that we can represent binary using polynomial. For example, $11$ can be represented as $x+1$. So when we compute $11\times11$, we should obtain $1001$, which is equal to $9$ in decimal. But if I use polynomial representation to compute, I obtain $(x+1)(x+1)=x^2+1$, which is $101$ in binary. Clearly it is not $9$ in decimal. Can anyone explain to me where I got wrong.
Remark: Another question: If we use different irreducible polynomial to compute the multiplication table of a Galois Field, say $GF(2^4)$, we will get different multiplication table for sure. I know the tables are isomorphic under with respect to multiplication. But would it affect computation if I use different irrducible polynomial in computer?
 A: Is $9+1=0$ in decimal? No, but modulo 10 it does.
That is where you have a problem here is that you aren't carrying over the term in the polynomial as if you treat the $2x$ as making another $x^2$ term, so you then have $2x^2$ for a moment, this then generates an $x^3$ term which would be the representation you seek. The question is whether $2=10$ or does $2=0$ in your binary representation?
Consider the simple example of $1 + 1$. Does this equal $x$ in your polynomial representation or $0$? I'd look at this as how do you handle numerical overflow?
If you take $11 * 11 = 121$  and then apply modulo 2 on each digit, you would get 101 which is what you have. The lack of carrying is what you have as a problem here, IMO.
To take this one step further. Consider if you could evaluate "121" using powers of 2 for each digit: $4+ 2*2 +1 = 9$. However, if you take out the middle term, this will leave you with the troubled answer you have. If the polynomial representation is to have the same positional notation then some things may have to be redefined, I believe.
