Let G(x) be the generator polynomial.

C(x) be the sent codeword, R(x) be the received code and E(x) be the error polynomial.

We can write: **R(x) = C(x) + E(x) ** { This can be written by mod 2 addition and skipping the carry}

As we know that: "**To detect error, G(x) should not divide R(x)**".

=> G(x) should not divide E(x). As, $\frac{R(x)}{G(x)} = \frac{C(x)}{G(x)} + \frac{E(x)}{G(x)}$

Further, $\frac{R(x)}{G(x)} = 0 + \frac{E(x)}{G(x)}$

This means if E(x) is not divisible by G(x) then R(x) will not also be divisible by G(x).

Now, **To detect odd number of bits in error, E(x) will contain odd number of terms. **

**Example: **let sent codeword= 101010 and Received codeword = 100100 i.e three bits (1st, 2nd and 3rd bit) are in error.

Therefore, $E(x) = x^{3}+x^{2}+x$.

**A) **G(x) contain more than 2 terms.

let, $G(x) = x^{3}+x^{2}+x$ then $\frac{E(x)}{G(x)} = \frac{x^{3}+x^{2}+x}{x^{3}+x^{2}+x}$

So, E(x) is divisible by G(x). So, A is not the right option.

**B) **G(x) does not divide by $1+x^{k}$, for any k not exceeding frame length.

here, k=6 [ frame length or code length ]

let $G(x) = x$ & $E(x) = x^{3}+x^{2}+x$.

$\frac{E(x)}{G(x)} = \frac{x^{3}+x^{2}+x}{x}$. here, E(x) can be divisible by G(x). So, B is not the right option.

**C) **$(1+x)$ is a factor of G(x).

$\frac{E(x)}{G(x)} = \frac{x^{3}+x^{2}+x}{(1+x)G(x)} = \frac{odd no. of terms}{even no. of terms}$

here, E(x) wil never be divisible by G(x). So, **(C) is correct option**.

**D) **G(x) has odd number of terms

let $G(x) = x^{3}+x^{2}+x$.

$\frac{E(x)}{G(x)} = \frac{odd No of terms}{Odd number of terms}$

here, G(x) may divide E(x). So, this canot be the option.

Hence, **Correct Ans: (C)**